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Time ago, I was asked during an interview to sum two Integers represented by arrays, putting the solution into another array. Part of the interview idea was for me to provide an efficient solution. Since then, I have been searching for a simple and efficient solution to this problem, and I didn't find none yet.
So I would like to share my solution with the community and ask if any of you can help to improve efficiency. This example looks like O(mn) where mn is the size of the biggest array between m or n and, where m and n represents the size of each integer array to sum. Thus, it looks as though it is working in linear time.
public class ArrayRudiments {
/**
* Add two integers represented by two arrays
*
* NOTE: Integer as a natural number, not as a programming language data type.
* Thus, the integer can be as big as the array permits.
*
* #param m first number as array
* #param n second number as array
* #return integer sum solution as array
*/
public Integer[] sum(Integer m[], Integer n[]) {
int carry = 0, csum = 0;
final Vector<Integer> solution = new Vector<Integer>();
for (int msize = m.length - 1, nsize = n.length - 1; msize >= 0 || nsize >= 0; msize--, nsize--) {
csum = (msize < 0 ? 0 : m[msize]) + (nsize < 0 ? 0 : n[nsize]) + carry;
carry = csum / 10;
solution.insertElementAt(csum % 10, 0);
}
if (carry > 0) {
solution.insertElementAt(carry, 0);
}
return solution.toArray(new Integer[] {});
}
}
The problem or trick here is that the not linear time job is carried out by the Vector class inserting new elements and resizing the solution array. This way, the solution is not working in linear time. Is it possible to create a similar solution without Vector class?
You can also see a working tested version of this code in https://github.com/lalounam/rudiments
As SSP has said in the comments, you should create an ArrayList with an initial capacity, of Math.max(m.length, n.length) + 1. That is the maximum number of digits of the sum.
int arrayCapacity = Math.max(m.length, n.length) + 1;
final List<Integer> solution = new ArrayList<>(arrayCapacity);
Then you need to fill this with 0s:
for (int i = 0 ; i < arrayCapacity ; i++) {
solution.add(0);
}
Here's the "trick". In the main for loop, you don't fill the array from the start, you fill it from the end. Instead of insertElementAt the start, you set the element at whatever index we are iterating at, plus 1, because solution is one longer than the longer of m and n.
solution.set(Math.max(msize, nsize) + 1, csum % 10);
This is essentially filling the list "from the back".
And at the end you resize the list like this:
if (carry > 0) {
solution.set(0, carry);
return solution.toArray(new Integer[] {});
} else {
return solution.subList(1, solution.size()).toArray(new Integer[] {});
}
excuse me for the confusing title, I need to implement an algorithm which can be simplified as the following:
given an array of integers, and the number of merges needed (denoted as k), return the maximum min value of the merged array, a merge can only happen with adjacent elements.
E.g. array = [3,2,8,2,9], k = 2
The maximum min value of the array after two merges is 5, and the merged array is [5, 10, 9]. In this example, we need to merge elements 3 & 2 and 8 & 2.
Any other merge strategies will yield min val that is smaller or equal to 5, e.g.:
[5,8,11], [3,10,11], [13,2,9](merged elements can be merged again)
What is the best data structure to represent the data and what is an efficient algorithm to address the problem? As far as I can think of, a greedy algorithm needs to be applied where a merge needs to happen with the current min value of the array and one of its smaller neighboring element.
Edit: I just realized that greedy algorithm might not apply, sorry for the misleading comment, if it doesn't distinguish between merging with left or right elements, this will generate the wrong answer. Take this as an example, given an array = [4,5,3,5], and we need to remove 2 elements.
With greedy, [4,5,3,5] -> [4,8,5] -> [12,5], so the answer is 5; however the correct answer should be 8 with the following merge sequence:
[4,5,3,5] -> [4,5,8] -> [9,8]
ValPosFrom is a simple class that stores those things, from being the place to merge from . you can get non deterministic results from things like List = 3,2,6,3,2 and k=1 it will merge one of the 2 mins to 5 but it doesn't matter which one. it converges when all of any positions neighbors values are unique.
private static List<Integer> merge(List<Integer> things, int merges) {
List<Integer> result = new ArrayList<>(things);
for (int i = 0; i < merges; i++) {
int min = Integer.MAX_VALUE;
List<Integer> positions = new ArrayList<>();
for (int j = 0; j < result.size(); j++) {
if (result.get(j) < min) {
positions.clear();
positions.add(j);
min = result.get(j);
} else if (result.get(j) == min) {
positions.add(j);
}
}
List<ValPosFrom> neighbors = new ArrayList<>();
positions.forEach(p -> {
if (p - 1 >= 0) {
neighbors.add(new ValPosFrom(result.get(p - 1), p - 1, p));
}
if (p + 1 < result.size()) {
neighbors.add(new ValPosFrom(result.get(p + 1), p + 1, p));
}
});
ValPosFrom vpf = Collections.min(neighbors, Comparator.comparingInt(v -> v.val));
result.set(vpf.pos, result.get(vpf.pos) + result.get(vpf.from));
result.remove(vpf.from);
}
return result;
}
In this case, the MAX is only 5, so I could check the duplicates one by one, but how could I do this in a simpler way? For example, what if the MAX has a value of 20?
Thanks.
int MAX = 5;
for (i = 1 , i <= MAX; i++)
{
drawNum[1] = (int)(Math.random()*MAX)+1;
while (drawNum[2] == drawNum[1])
{
drawNum[2] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[3] == drawNum[1]) || (drawNum[3] == drawNum[2]) )
{
drawNum[3] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[4] == drawNum[1]) || (drawNum[4] == drawNum[2]) || (drawNum[4] == drawNum[3]) )
{
drawNum[4] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[5] == drawNum[1]) ||
(drawNum[5] == drawNum[2]) ||
(drawNum[5] == drawNum[3]) ||
(drawNum[5] == drawNum[4]) )
{
drawNum[5] = (int)(Math.random()*MAX)+1;
}
}
The simplest way would be to create a list of the possible numbers (1..20 or whatever) and then shuffle them with Collections.shuffle. Then just take however many elements you want. This is great if your range is equal to the number of elements you need in the end (e.g. for shuffling a deck of cards).
That doesn't work so well if you want (say) 10 random elements in the range 1..10,000 - you'd end up doing a lot of work unnecessarily. At that point, it's probably better to keep a set of values you've generated so far, and just keep generating numbers in a loop until the next one isn't already present:
if (max < numbersNeeded)
{
throw new IllegalArgumentException("Can't ask for more numbers than are available");
}
Random rng = new Random(); // Ideally just create one instance globally
// Note: use LinkedHashSet to maintain insertion order
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
Integer next = rng.nextInt(max) + 1;
// As we're adding to a set, this will automatically do a containment check
generated.add(next);
}
Be careful with the set choice though - I've very deliberately used LinkedHashSet as it maintains insertion order, which we care about here.
Yet another option is to always make progress, by reducing the range each time and compensating for existing values. So for example, suppose you wanted 3 values in the range 0..9. On the first iteration you'd generate any number in the range 0..9 - let's say you generate a 4.
On the second iteration you'd then generate a number in the range 0..8. If the generated number is less than 4, you'd keep it as is... otherwise you add one to it. That gets you a result range of 0..9 without 4. Suppose we get 7 that way.
On the third iteration you'd generate a number in the range 0..7. If the generated number is less than 4, you'd keep it as is. If it's 4 or 5, you'd add one. If it's 6 or 7, you'd add two. That way the result range is 0..9 without 4 or 6.
Here's how I'd do it
import java.util.ArrayList;
import java.util.Random;
public class Test {
public static void main(String[] args) {
int size = 20;
ArrayList<Integer> list = new ArrayList<Integer>(size);
for(int i = 1; i <= size; i++) {
list.add(i);
}
Random rand = new Random();
while(list.size() > 0) {
int index = rand.nextInt(list.size());
System.out.println("Selected: "+list.remove(index));
}
}
}
As the esteemed Mr Skeet has pointed out:
If n is the number of randomly selected numbers you wish to choose and N is the total sample space of numbers available for selection:
If n << N, you should just store the numbers that you have picked and check a list to see if the number selected is in it.
If n ~= N, you should probably use my method, by populating a list containing the entire sample space and then removing numbers from it as you select them.
//random numbers are 0,1,2,3
ArrayList<Integer> numbers = new ArrayList<Integer>();
Random randomGenerator = new Random();
while (numbers.size() < 4) {
int random = randomGenerator .nextInt(4);
if (!numbers.contains(random)) {
numbers.add(random);
}
}
This would be a lot simpler in java-8:
Stream.generate(new Random()::ints)
.flatMap(IntStream::boxed)
.distinct()
.limit(16) // whatever limit you might need
.toArray(Integer[]::new);
There is another way of doing "random" ordered numbers with LFSR, take a look at:
http://en.wikipedia.org/wiki/Linear_feedback_shift_register
with this technique you can achieve the ordered random number by index and making sure the values are not duplicated.
But these are not TRUE random numbers because the random generation is deterministic.
But depending your case you can use this technique reducing the amount of processing on random number generation when using shuffling.
Here a LFSR algorithm in java, (I took it somewhere I don't remeber):
public final class LFSR {
private static final int M = 15;
// hard-coded for 15-bits
private static final int[] TAPS = {14, 15};
private final boolean[] bits = new boolean[M + 1];
public LFSR() {
this((int)System.currentTimeMillis());
}
public LFSR(int seed) {
for(int i = 0; i < M; i++) {
bits[i] = (((1 << i) & seed) >>> i) == 1;
}
}
/* generate a random int uniformly on the interval [-2^31 + 1, 2^31 - 1] */
public short nextShort() {
//printBits();
// calculate the integer value from the registers
short next = 0;
for(int i = 0; i < M; i++) {
next |= (bits[i] ? 1 : 0) << i;
}
// allow for zero without allowing for -2^31
if (next < 0) next++;
// calculate the last register from all the preceding
bits[M] = false;
for(int i = 0; i < TAPS.length; i++) {
bits[M] ^= bits[M - TAPS[i]];
}
// shift all the registers
for(int i = 0; i < M; i++) {
bits[i] = bits[i + 1];
}
return next;
}
/** returns random double uniformly over [0, 1) */
public double nextDouble() {
return ((nextShort() / (Integer.MAX_VALUE + 1.0)) + 1.0) / 2.0;
}
/** returns random boolean */
public boolean nextBoolean() {
return nextShort() >= 0;
}
public void printBits() {
System.out.print(bits[M] ? 1 : 0);
System.out.print(" -> ");
for(int i = M - 1; i >= 0; i--) {
System.out.print(bits[i] ? 1 : 0);
}
System.out.println();
}
public static void main(String[] args) {
LFSR rng = new LFSR();
Vector<Short> vec = new Vector<Short>();
for(int i = 0; i <= 32766; i++) {
short next = rng.nextShort();
// just testing/asserting to make
// sure the number doesn't repeat on a given list
if (vec.contains(next))
throw new RuntimeException("Index repeat: " + i);
vec.add(next);
System.out.println(next);
}
}
}
Another approach which allows you to specify how many numbers you want with size and the min and max values of the returned numbers
public static int getRandomInt(int min, int max) {
Random random = new Random();
return random.nextInt((max - min) + 1) + min;
}
public static ArrayList<Integer> getRandomNonRepeatingIntegers(int size, int min,
int max) {
ArrayList<Integer> numbers = new ArrayList<Integer>();
while (numbers.size() < size) {
int random = getRandomInt(min, max);
if (!numbers.contains(random)) {
numbers.add(random);
}
}
return numbers;
}
To use it returning 7 numbers between 0 and 25.
ArrayList<Integer> list = getRandomNonRepeatingIntegers(7, 0, 25);
for (int i = 0; i < list.size(); i++) {
System.out.println("" + list.get(i));
}
The most efficient, basic way to have non-repeating random numbers is explained by this pseudo-code. There is no need to have nested loops or hashed lookups:
// get 5 unique random numbers, possible values 0 - 19
// (assume desired number of selections < number of choices)
const int POOL_SIZE = 20;
const int VAL_COUNT = 5;
declare Array mapping[POOL_SIZE];
declare Array results[VAL_COUNT];
declare i int;
declare r int;
declare max_rand int;
// create mapping array
for (i=0; i<POOL_SIZE; i++) {
mapping[i] = i;
}
max_rand = POOL_SIZE-1; // start loop searching for maximum value (19)
for (i=0; i<VAL_COUNT; i++) {
r = Random(0, max_rand); // get random number
results[i] = mapping[r]; // grab number from map array
mapping[r] = max_rand; // place item past range at selected location
max_rand = max_rand - 1; // reduce random scope by 1
}
Suppose first iteration generated random number 3 to start (from 0 - 19). This would make results[0] = mapping[3], i.e., the value 3. We'd then assign mapping[3] to 19.
In the next iteration, the random number was 5 (from 0 - 18). This would make results[1] = mapping[5], i.e., the value 5. We'd then assign mapping[5] to 18.
Now suppose the next iteration chose 3 again (from 0 - 17). results[2] would be assigned the value of mapping[3], but now, this value is not 3, but 19.
This same protection persists for all numbers, even if you got the same number 5 times in a row. E.g., if the random number generator gave you 0 five times in a row, the results would be: [ 0, 19, 18, 17, 16 ].
You would never get the same number twice.
Generating all the indices of a sequence is generally a bad idea, as it might take a lot of time, especially if the ratio of the numbers to be chosen to MAX is low (the complexity becomes dominated by O(MAX)). This gets worse if the ratio of the numbers to be chosen to MAX approaches one, as then removing the chosen indices from the sequence of all also becomes expensive (we approach O(MAX^2/2)). But for small numbers, this generally works well and is not particularly error-prone.
Filtering the generated indices by using a collection is also a bad idea, as some time is spent in inserting the indices into the sequence, and progress is not guaranteed as the same random number can be drawn several times (but for large enough MAX it is unlikely). This could be close to complexity O(k n log^2(n)/2), ignoring the duplicates and assuming the collection uses a tree for efficient lookup (but with a significant constant cost k of allocating the tree nodes and possibly having to rebalance).
Another option is to generate the random values uniquely from the beginning, guaranteeing progress is being made. That means in the first round, a random index in [0, MAX] is generated:
items i0 i1 i2 i3 i4 i5 i6 (total 7 items)
idx 0 ^^ (index 2)
In the second round, only [0, MAX - 1] is generated (as one item was already selected):
items i0 i1 i3 i4 i5 i6 (total 6 items)
idx 1 ^^ (index 2 out of these 6, but 3 out of the original 7)
The values of the indices then need to be adjusted: if the second index falls in the second half of the sequence (after the first index), it needs to be incremented to account for the gap. We can implement this as a loop, allowing us to select arbitrary number of unique items.
For short sequences, this is quite fast O(n^2/2) algorithm:
void RandomUniqueSequence(std::vector<int> &rand_num,
const size_t n_select_num, const size_t n_item_num)
{
assert(n_select_num <= n_item_num);
rand_num.clear(); // !!
// b1: 3187.000 msec (the fastest)
// b2: 3734.000 msec
for(size_t i = 0; i < n_select_num; ++ i) {
int n = n_Rand(n_item_num - i - 1);
// get a random number
size_t n_where = i;
for(size_t j = 0; j < i; ++ j) {
if(n + j < rand_num[j]) {
n_where = j;
break;
}
}
// see where it should be inserted
rand_num.insert(rand_num.begin() + n_where, 1, n + n_where);
// insert it in the list, maintain a sorted sequence
}
// tier 1 - use comparison with offset instead of increment
}
Where n_select_num is your 5 and n_number_num is your MAX. The n_Rand(x) returns random integers in [0, x] (inclusive). This can be made a bit faster if selecting a lot of items (e.g. not 5 but 500) by using binary search to find the insertion point. To do that, we need to make sure that we meet the requirements.
We will do binary search with the comparison n + j < rand_num[j] which is the same as n < rand_num[j] - j. We need to show that rand_num[j] - j is still a sorted sequence for a sorted sequence rand_num[j]. This is fortunately easily shown, as the lowest distance between two elements of the original rand_num is one (the generated numbers are unique, so there is always difference of at least 1). At the same time, if we subtract the indices j from all the elements rand_num[j], the differences in index are exactly 1. So in the "worst" case, we get a constant sequence - but never decreasing. The binary search can therefore be used, yielding O(n log(n)) algorithm:
struct TNeedle { // in the comparison operator we need to make clear which argument is the needle and which is already in the list; we do that using the type system.
int n;
TNeedle(int _n)
:n(_n)
{}
};
class CCompareWithOffset { // custom comparison "n < rand_num[j] - j"
protected:
std::vector<int>::iterator m_p_begin_it;
public:
CCompareWithOffset(std::vector<int>::iterator p_begin_it)
:m_p_begin_it(p_begin_it)
{}
bool operator ()(const int &r_value, TNeedle n) const
{
size_t n_index = &r_value - &*m_p_begin_it;
// calculate index in the array
return r_value < n.n + n_index; // or r_value - n_index < n.n
}
bool operator ()(TNeedle n, const int &r_value) const
{
size_t n_index = &r_value - &*m_p_begin_it;
// calculate index in the array
return n.n + n_index < r_value; // or n.n < r_value - n_index
}
};
And finally:
void RandomUniqueSequence(std::vector<int> &rand_num,
const size_t n_select_num, const size_t n_item_num)
{
assert(n_select_num <= n_item_num);
rand_num.clear(); // !!
// b1: 3578.000 msec
// b2: 1703.000 msec (the fastest)
for(size_t i = 0; i < n_select_num; ++ i) {
int n = n_Rand(n_item_num - i - 1);
// get a random number
std::vector<int>::iterator p_where_it = std::upper_bound(rand_num.begin(), rand_num.end(),
TNeedle(n), CCompareWithOffset(rand_num.begin()));
// see where it should be inserted
rand_num.insert(p_where_it, 1, n + p_where_it - rand_num.begin());
// insert it in the list, maintain a sorted sequence
}
// tier 4 - use binary search
}
I have tested this on three benchmarks. First, 3 numbers were chosen out of 7 items, and a histogram of the items chosen was accumulated over 10,000 runs:
4265 4229 4351 4267 4267 4364 4257
This shows that each of the 7 items was chosen approximately the same number of times, and there is no apparent bias caused by the algorithm. All the sequences were also checked for correctness (uniqueness of contents).
The second benchmark involved choosing 7 numbers out of 5000 items. The time of several versions of the algorithm was accumulated over 10,000,000 runs. The results are denoted in comments in the code as b1. The simple version of the algorithm is slightly faster.
The third benchmark involved choosing 700 numbers out of 5000 items. The time of several versions of the algorithm was again accumulated, this time over 10,000 runs. The results are denoted in comments in the code as b2. The binary search version of the algorithm is now more than two times faster than the simple one.
The second method starts being faster for choosing more than cca 75 items on my machine (note that the complexity of either algorithm does not depend on the number of items, MAX).
It is worth mentioning that the above algorithms generate the random numbers in ascending order. But it would be simple to add another array to which the numbers would be saved in the order in which they were generated, and returning that instead (at negligible additional cost O(n)). It is not necessary to shuffle the output: that would be much slower.
Note that the sources are in C++, I don't have Java on my machine, but the concept should be clear.
EDIT:
For amusement, I have also implemented the approach that generates a list with all the indices 0 .. MAX, chooses them randomly and removes them from the list to guarantee uniqueness. Since I've chosen quite high MAX (5000), the performance is catastrophic:
// b1: 519515.000 msec
// b2: 20312.000 msec
std::vector<int> all_numbers(n_item_num);
std::iota(all_numbers.begin(), all_numbers.end(), 0);
// generate all the numbers
for(size_t i = 0; i < n_number_num; ++ i) {
assert(all_numbers.size() == n_item_num - i);
int n = n_Rand(n_item_num - i - 1);
// get a random number
rand_num.push_back(all_numbers[n]); // put it in the output list
all_numbers.erase(all_numbers.begin() + n); // erase it from the input
}
// generate random numbers
I have also implemented the approach with a set (a C++ collection), which actually comes second on benchmark b2, being only about 50% slower than the approach with the binary search. That is understandable, as the set uses a binary tree, where the insertion cost is similar to binary search. The only difference is the chance of getting duplicate items, which slows down the progress.
// b1: 20250.000 msec
// b2: 2296.000 msec
std::set<int> numbers;
while(numbers.size() < n_number_num)
numbers.insert(n_Rand(n_item_num - 1)); // might have duplicates here
// generate unique random numbers
rand_num.resize(numbers.size());
std::copy(numbers.begin(), numbers.end(), rand_num.begin());
// copy the numbers from a set to a vector
Full source code is here.
Your problem seems to reduce to choose k elements at random from a collection of n elements. The Collections.shuffle answer is thus correct, but as pointed out inefficient: its O(n).
Wikipedia: Fisher–Yates shuffle has a O(k) version when the array already exists. In your case, there is no array of elements and creating the array of elements could be very expensive, say if max were 10000000 instead of 20.
The shuffle algorithm involves initializing an array of size n where every element is equal to its index, picking k random numbers each number in a range with the max one less than the previous range, then swapping elements towards the end of the array.
You can do the same operation in O(k) time with a hashmap although I admit its kind of a pain. Note that this is only worthwhile if k is much less than n. (ie k ~ lg(n) or so), otherwise you should use the shuffle directly.
You will use your hashmap as an efficient representation of the backing array in the shuffle algorithm. Any element of the array that is equal to its index need not appear in the map. This allows you to represent an array of size n in constant time, there is no time spent initializing it.
Pick k random numbers: the first is in the range 0 to n-1, the second 0 to n-2, the third 0 to n-3 and so on, thru n-k.
Treat your random numbers as a set of swaps. The first random index swaps to the final position. The second random index swaps to the second to last position. However, instead of working against a backing array, work against your hashmap. Your hashmap will store every item that is out of position.
int getValue(i)
{
if (map.contains(i))
return map[i];
return i;
}
void setValue(i, val)
{
if (i == val)
map.remove(i);
else
map[i] = val;
}
int[] chooseK(int n, int k)
{
for (int i = 0; i < k; i++)
{
int randomIndex = nextRandom(0, n - i); //(n - i is exclusive)
int desiredIndex = n-i-1;
int valAtRandom = getValue(randomIndex);
int valAtDesired = getValue(desiredIndex);
setValue(desiredIndex, valAtRandom);
setValue(randomIndex, valAtDesired);
}
int[] output = new int[k];
for (int i = 0; i < k; i++)
{
output[i] = (getValue(n-i-1));
}
return output;
}
You could use one of the classes implementing the Set interface (API), and then each number you generate, use Set.add() to insert it.
If the return value is false, you know the number has already been generated before.
Instead of doing all this create a LinkedHashSet object and random numbers to it by Math.random() function .... if any duplicated entry occurs the LinkedHashSet object won't add that number to its List ... Since in this Collection Class no duplicate values are allowed .. in the end u get a list of random numbers having no duplicated values .... :D
With Java 8 upwards you can use the ints method from the IntStream interface:
Returns an effectively unlimited stream of pseudorandom int values.
Random r = new Random();
int randomNumberOrigin = 0;
int randomNumberBound = 10;
int size = 5;
int[] unique = r.ints(randomNumberOrigin, randomNumberBound)
.distinct()
.limit(size)
.toArray();
Following code create a sequence random number between [1,m] that was not generated before.
public class NewClass {
public List<Integer> keys = new ArrayList<Integer>();
public int rand(int m) {
int n = (int) (Math.random() * m + 1);
if (!keys.contains(n)) {
keys.add(n);
return n;
} else {
return rand(m);
}
}
public static void main(String[] args) {
int m = 4;
NewClass ne = new NewClass();
for (int i = 0; i < 4; i++) {
System.out.println(ne.rand(m));
}
System.out.println("list: " + ne.keys);
}
}
The most easy way is use nano DateTime as long format.
System.nanoTime();
There is algorithm of card batch: you create ordered array of numbers (the "card batch") and in every iteration you select a number at random position from it (removing the selected number from the "card batch" of course).
Here is an efficient solution for fast creation of a randomized array. After randomization you can simply pick the n-th element e of the array, increment n and return e. This solution has O(1) for getting a random number and O(n) for initialization, but as a tradeoff requires a good amount of memory if n gets large enough.
There is a more efficient and less cumbersome solution for integers than a Collections.shuffle.
The problem is the same as successively picking items from only the un-picked items in a set and setting them in order somewhere else. This is exactly like randomly dealing cards or drawing winning raffle tickets from a hat or bin.
This algorithm works for loading any array and achieving a random order at the end of the load. It also works for adding into a List collection (or any other indexed collection) and achieving a random sequence in the collection at the end of the adds.
It can be done with a single array, created once, or a numerically ordered collectio, such as a List, in place. For an array, the initial array size needs to be the exact size to contain all the intended values. If you don't know how many values might occur in advance, using a numerically orderred collection, such as an ArrayList or List, where the size is not immutable, will also work. It will work universally for an array of any size up to Integer.MAX_VALUE which is just over 2,000,000,000. List objects will have the same index limits. Your machine may run out of memory before you get to an array of that size. It may be more efficient to load an array typed to the object types and convert it to some collection, after loading the array. This is especially true if the target collection is not numerically indexed.
This algorithm, exactly as written, will create a very even distribution where there are no duplicates. One aspect that is VERY IMPORTANT is that it has to be possible for the insertion of the next item to occur up to the current size + 1. Thus, for the second item, it could be possible to store it in location 0 or location 1. For the 20th item, it could be possible to store it in any location, 0 through 19. It is just as possible the first item to stay in location 0 as it is for it to end up in any other location. It is just as possible for the next new item to go anywhere, including the next new location.
The randomness of the sequence will be as random as the randomness of the random number generator.
This algorithm can also be used to load reference types into random locations in an array. Since this works with an array, it can also work with collections. That means you don't have to create the collection and then shuffle it or have it ordered on whatever orders the objects being inserted. The collection need only have the ability to insert an item anywhere in the collection or append it.
// RandomSequence.java
import java.util.Random;
public class RandomSequence {
public static void main(String[] args) {
// create an array of the size and type for which
// you want a random sequence
int[] randomSequence = new int[20];
Random randomNumbers = new Random();
for (int i = 0; i < randomSequence.length; i++ ) {
if (i == 0) { // seed first entry in array with item 0
randomSequence[i] = 0;
} else { // for all other items...
// choose a random pointer to the segment of the
// array already containing items
int pointer = randomNumbers.nextInt(i + 1);
randomSequence[i] = randomSequence[pointer];
randomSequence[pointer] = i;
// note that if pointer & i are equal
// the new value will just go into location i and possibly stay there
// this is VERY IMPORTANT to ensure the sequence is really random
// and not biased
} // end if...else
} // end for
for (int number: randomSequence) {
System.out.printf("%2d ", number);
} // end for
} // end main
} // end class RandomSequence
It really all depends on exactly WHAT you need the random generation for, but here's my take.
First, create a standalone method for generating the random number.
Be sure to allow for limits.
public static int newRandom(int limit){
return generatedRandom.nextInt(limit); }
Next, you will want to create a very simple decision structure that compares values. This can be done in one of two ways. If you have a very limited amount of numbers to verify, a simple IF statement will suffice:
public static int testDuplicates(int int1, int int2, int int3, int int4, int int5){
boolean loopFlag = true;
while(loopFlag == true){
if(int1 == int2 || int1 == int3 || int1 == int4 || int1 == int5 || int1 == 0){
int1 = newRandom(75);
loopFlag = true; }
else{
loopFlag = false; }}
return int1; }
The above compares int1 to int2 through int5, as well as making sure that there are no zeroes in the randoms.
With these two methods in place, we can do the following:
num1 = newRandom(limit1);
num2 = newRandom(limit1);
num3 = newRandom(limit1);
num4 = newRandom(limit1);
num5 = newRandom(limit1);
Followed By:
num1 = testDuplicates(num1, num2, num3, num4, num5);
num2 = testDuplicates(num2, num1, num3, num4, num5);
num3 = testDuplicates(num3, num1, num2, num4, num5);
num4 = testDuplicates(num4, num1, num2, num3, num5);
num5 = testDuplicates(num5, num1, num2, num3, num5);
If you have a longer list to verify, then a more complex method will yield better results both in clarity of code and in processing resources.
Hope this helps. This site has helped me so much, I felt obliged to at least TRY to help as well.
I created a snippet that generates no duplicate random integer. the advantage of this snippet is that you can assign the list of an array to it and generate the random item, too.
No duplication random generator class
With Java 8 using the below code, you can create 10 distinct random Integer Numbers within a range of 1000.
Random random = new Random();
Integer[] input9 = IntStream.range(1, 10).map(i -> random.nextInt(1000)).boxed().distinct()
.toArray(Integer[]::new);
System.out.println(Arrays.toString(input9));
Modify the range to generate more numbers example : range(1,X). It will generate X distinct random numbers.
Modify the nextInt value to select the random number range : random.nextInt(Y)::random number will be generated within the range Y
I have been asked this question in a job interview and I have been wondering about the right answer.
You have an array of numbers from 0 to n-1, one of the numbers is removed, and replaced with a number already in the array which makes a duplicate of that number. How can we detect this duplicate in time O(n)?
For example, an array of 4,1,2,3 would become 4,1,2,2.
The easy solution of time O(n2) is to use a nested loop to look for the duplicate of each element.
This can be done in O(n) time and O(1) space.
(The algorithm only works because the numbers are consecutive integers in a known range):
In a single pass through the vector, compute the sum of all the numbers, and the sum of the squares of all the numbers.
Subtract the sum of all the numbers from N(N-1)/2. Call this A.
Subtract the sum of the squares from N(N-1)(2N-1)/6. Divide this by A. Call the result B.
The number which was removed is (B + A)/2 and the number it was replaced with is (B - A)/2.
Example:
The vector is [0, 1, 1, 2, 3, 5]:
N = 6
Sum of the vector is 0 + 1 + 1 + 2 + 3 + 5 = 12. N(N-1)/2 is 15. A = 3.
Sum of the squares is 0 + 1 + 1 + 4 + 9 + 25 = 40. N(N-1)(2N-1)/6 is 55. B = (55 - 40)/A = 5.
The number which was removed is (5 + 3) / 2 = 4.
The number it was replaced by is (5 - 3) / 2 = 1.
Why it works:
The sum of the original vector [0, ..., N-1] is N(N-1)/2. Suppose the value a was removed and replaced by b. Now the sum of the modified vector will be N(N-1)/2 + b - a. If we subtract the sum of the modified vector from N(N-1)/2 we get a - b. So A = a - b.
Similarly, the sum of the squares of the original vector is N(N-1)(2N-1)/6. The sum of the squares of the modified vector is N(N-1)(2N-1)/6 + b2 - a2. Subtracting the sum of the squares of the modified vector from the original sum gives a2 - b2, which is the same as (a+b)(a-b). So if we divide it by a - b (i.e., A), we get B = a + b.
Now B + A = a + b + a - b = 2a and B - A = a + b - (a - b) = 2b.
We have the original array int A[N]; Create a second array bool B[N] too, of type bool=false. Iterate the first array and set B[A[i]]=true if was false, else bing!
You can do it in O(N) time without any extra space. Here is how the algorithm works :
Iterate through array in the following manner :
For each element encountered, set its corresponding index value to negative.
Eg : if you find a[0] = 2. Got to a[2] and negate the value.
By doing this you flag it to be encountered. Since you know you cannot have negative numbers, you also know that you are the one who negated it.
Check if index corresponding to the value is already flagged negative, if yes you get the duplicated element. Eg : if a[0]=2 , go to a[2] and check if it is negative.
Lets say you have following array :
int a[] = {2,1,2,3,4};
After first element your array will be :
int a[] = {2,1,-2,3,4};
After second element your array will be :
int a[] = {2,-1,-2,3,4};
When you reach third element you go to a[2] and see its already negative. You get the duplicate.
Scan the array 3 times:
XOR together all the array elements -> A. XOR together all the numbers from 0 to N-1 -> B. Now A XOR B = X XOR D, where X is the removed element, and D is the duplicate element.
Choose any non-zero bit in A XOR B. XOR together all the array elements where this bit is set -> A1. XOR together all the numbers from 0 to N-1 where this bit is set -> B1. Now either A1 XOR B1 = X or A1 XOR B1 = D.
Scan the array once more and try to find A1 XOR B1. If it is found, this is the duplicate element. If not, the duplicate element is A XOR B XOR A1 XOR B1.
Use a HashSet to hold all numbers already seen. It operates in (amortized) O(1) time, so the total is O(N).
I suggest using a BitSet. We know N is small enough for array indexing, so the BitSet will be of reasonable size.
For each element of the array, check the bit corresponding to its value. If it is already set, that is the duplicate. If not, set the bit.
#rici is right about the time and space usage: "This can be done in O(n) time and O(1) space."
However, the question can be expanded to broader requirement: it's not necessary that there is only one duplicate number, and numbers might not be consecutive.
OJ puts it this way here:
(note 3 apparently can be narrowed)
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
The question is very well explained and answered here by Keith Schwarz, using Floyd's cycle-finding algorithm:
The main trick we need to use to solve this problem is to notice that because we have an array of n elements ranging from 0 to n - 2, we can think of the array as defining a function f from the set {0, 1, ..., n - 1} onto itself. This function is defined by f(i) = A[i]. Given this setup, a duplicated value corresponds to a pair of indices i != j such that f(i) = f(j). Our challenge, therefore, is to find this pair (i, j). Once we have it, we can easily find the duplicated value by just picking f(i) = A[i].
But how are we to find this repeated value? It turns out that this is a well-studied problem in computer science called cycle detection. The general form of the problem is as follows. We are given a function f. Define the sequence x_i as
x_0 = k (for some k)
x_1 = f(x_0)
x_2 = f(f(x_0))
...
x_{n+1} = f(x_n)
Assuming that f maps from a domain into itself, this function will have one of three forms. First, if the domain is infinite, then the sequence could be infinitely long and nonrepeating. For example, the function f(n) = n + 1 on the integers has this property - no number is ever duplicated. Second, the sequence could be a closed loop, which means that there is some i so that x_0 = x_i. In this case, the sequence cycles through some fixed set of values indefinitely. Finally, the sequence could be "rho-shaped." In this case, the sequence looks something like this:
x_0 -> x_1 -> ... x_k -> x_{k+1} ... -> x_{k+j}
^ |
| |
+-----------------------+
That is, the sequence begins with a chain of elements that enters a cycle, then cycles around indefinitely. We'll denote the first element of the cycle that is reached in the sequence the "entry" of the cycle.
An python implementation can also be found here:
def findDuplicate(self, nums):
# The "tortoise and hare" step. We start at the end of the array and try
# to find an intersection point in the cycle.
slow = 0
fast = 0
# Keep advancing 'slow' by one step and 'fast' by two steps until they
# meet inside the loop.
while True:
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast:
break
# Start up another pointer from the end of the array and march it forward
# until it hits the pointer inside the array.
finder = 0
while True:
slow = nums[slow]
finder = nums[finder]
# If the two hit, the intersection index is the duplicate element.
if slow == finder:
return slow
Use hashtable. Including an element in a hashtable is O(1).
One working solution:
asume number are integers
create an array of [0 .. N]
int[] counter = new int[N];
Then iterate read and increment the counter:
if (counter[val] >0) {
// duplicate
} else {
counter[val]++;
}
This can be done in O(n) time and O(1) space.
Without modifying the input array
The idea is similar to finding the starting node of a loop in a linked list.
Maintain two pointers: fast and slow
slow = a[0]
fast = a[a[0]]
loop till slow != fast
Once we find the loop (slow == fast)
Reset slow back to zero
slow = 0
find the starting node
while(slow != fast){
slow = a[slow];
fast = a[fast];
}
slow is your duplicate number.
Here's a Java implementation:
class Solution {
public int findDuplicate(int[] nums) {
if(nums.length <= 1) return -1;
int slow = nums[0], fast = nums[nums[0]]; //slow = head.next, fast = head.next.next
while(slow != fast){ //check for loop
slow = nums[slow];
fast = nums[nums[fast]];
}
if(slow != fast) return -1;
slow = 0; //reset one pointer
while(slow != fast){ //find starting point of loop
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
}
This is an alternative solution in O(n) time and O(1) space. It is similar to rici's. I find it a bit easier to understand but, in practice, it will overflow faster.
Let X be the missing number and R be the repeated number.
We can assume the numbers are from [1..n], i.e. zero does not appear. In fact, while looping through the array, we can test if zero was found and return immediately if not.
Now consider:
sum(A) = n (n + 1) / 2 - X + R
product(A) = n! R / X
where product(A) is the product of all element in A skipping the zero. We have two equations in two unknowns from which X and R can be derived algebraically.
Edit: by popular demand, here is a worked-out example:
Let's set:
S = sum(A) - n (n + 1) / 2
P = n! / product(A)
Then our equations become:
R - X = S
X = R P
which can be solved to:
R = S / (1 - P)
X = P R = P S / (1 - P)
Example:
A = [0 1 2 2 4]
n = A.length - 1 = 4
S = (1 + 2 + 2 + 4) - 4 * 5 / 2 = -1
P = 4! / (1 * 2 * 2 * 4) = 3 / 2
R = -1 / (1 - 3/2) = -1 / -1/2 = 2
X = 3/2 * 2 = 3
You could proceed as follows:
sort your array by using a Linear-time sorting algorithm (e.g. Counting sort) - O(N)
scan the sorted array and stop as soon as two consecutive elements are equal - O(N)
public class FindDuplicate {
public static void main(String[] args) {
// assume the array is sorted, otherwise first we have to sort it.
// time efficiency is o(n)
int elementData[] = new int[] { 1, 2, 3, 3, 4, 5, 6, 8, 8 };
int count = 1;
int element1;
int element2;
for (int i = 0; i < elementData.length - 1; i++) {
element1 = elementData[i];
element2 = elementData[count];
count++;
if (element1 == element2) {
System.out.println(element2);
}
}
}
}
public void duplicateNumberInArray {
int a[] = new int[10];
Scanner inp = new Scanner(System.in);
for(int i=1;i<=5;i++){
System.out.println("enter no. ");
a[i] = inp.nextInt();
}
Set<Integer> st = new HashSet<Integer>();
Set<Integer> s = new HashSet<Integer>();
for(int i=1;i<=5;i++){
if(!st.add(a[i])){
s.add(a[i]);
}
}
Iterator<Integer> itr = s.iterator();
System.out.println("Duplicate numbers are");
while(itr.hasNext()){
System.out.println(itr.next());
}
}
First of all creating an array of integer using Scanner class. Then iterating a loop through the numbers and checking if the number can be added to set (Numbers can be added to set only when that particular number should not be in set already, means set does not allow duplicate no. to add and return a boolean vale FALSE on adding duplicate value).If no. cannot be added means it is duplicate so add that duplicate number into another set, so that we can print later. Please note onething that we are adding the duplicate number into a set because it might be possible that duplicate number might be repeated several times, hence add it only once.At last we are printing set using Iterator.
//This is similar to the HashSet approach but uses only one data structure:
int[] a = { 1, 4, 6, 7, 4, 6, 5, 22, 33, 44, 11, 5 };
LinkedHashMap<Integer, Integer> map = new LinkedHashMap<Integer, Integer>();
for (int i : a) {
map.put(i, map.containsKey(i) ? (map.get(i)) + 1 : 1);
}
Set<Entry<Integer, Integer>> es = map.entrySet();
Iterator<Entry<Integer, Integer>> it = es.iterator();
while (it.hasNext()) {
Entry<Integer, Integer> e = it.next();
if (e.getValue() > 1) {
System.out.println("Dupe " + e.getKey());
}
}
We can do using hashMap efficiently:
Integer[] a = {1,2,3,4,0,1,5,2,1,1,1,};
HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
for(int x : a)
{
if (map.containsKey(x)) map.put(x,map.get(x)+1);
else map.put(x,1);
}
Integer [] keys = map.keySet().toArray(new Integer[map.size()]);
for(int x : keys)
{
if(map.get(x)!=1)
{
System.out.println(x+" repeats : "+map.get(x));
}
}
This program is based on c# and if you want to do this program using another programming language you have to firstly change an array in accending order and compare the first element to the second element.If it is equal then repeated number found.Program is
int[] array=new int[]{1,2,3,4,5,6,7,8,9,4};
Array.Sort(array);
for(int a=0;a<array.Length-1;a++)
{
if(array[a]==array[a+1]
{
Console.WriteLine("This {0} element is repeated",array[a]);
}
}
Console.WriteLine("Not repeated number in array");
sort the array O(n ln n)
using the sliding window trick to traverse the array O(n)
Space is O(1)
Arrays.sort(input);
for(int i = 0, j = 1; j < input.length ; j++, i++){
if( input[i] == input[j]){
System.out.println(input[i]);
while(j < input.length && input[i] == input[j]) j++;
i = j - 1;
}
}
Test case int[] { 1, 2, 3, 7, 7, 8, 3, 5, 7, 1, 2, 7 }
output 1, 2, 3, 7
Traverse through the array and check the sign of array[abs(array[i])], if positive make it as negative and if it is negative then print it, as follows:
import static java.lang.Math.abs;
public class FindRepeatedNumber {
private static void findRepeatedNumber(int arr[]) {
int i;
for (i = 0; i < arr.length; i++) {
if (arr[abs(arr[i])] > 0)
arr[abs(arr[i])] = -arr[abs(arr[i])];
else {
System.out.print(abs(arr[i]) + ",");
}
}
}
public static void main(String[] args) {
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
findRepeatedNumber(arr);
}
}
Reference: http://www.geeksforgeeks.org/find-duplicates-in-on-time-and-constant-extra-space/
As described,
You have an array of numbers from 0 to n-1, one of the numbers is
removed, and replaced with a number already in the array which makes a
duplicate of that number.
I'm assuming elements in the array are sorted except the duplicate entry. If this is the scenario , we can achieve the goal easily as below :
public static void main(String[] args) {
//int arr[] = { 0, 1, 2, 2, 3 };
int arr[] = { 1, 2, 3, 4, 3, 6 };
int len = arr.length;
int iMax = arr[0];
for (int i = 1; i < len; i++) {
iMax = Math.max(iMax, arr[i]);
if (arr[i] < iMax) {
System.out.println(arr[i]);
break;
}else if(arr[i+1] <= iMax) {
System.out.println(arr[i+1]);
break;
}
}
}
O(n) time and O(1) space ;please share your thoughts.
Here is the simple solution with hashmap in O(n) time.
#include<iostream>
#include<map>
using namespace std;
int main()
{
int a[]={1,3,2,7,5,1,8,3,6,10};
map<int,int> mp;
for(int i=0;i<10;i++){
if(mp.find(a[i]) == mp.end())
mp.insert({a[i],1});
else
mp[a[i]]++;
}
for(auto i=mp.begin();i!=mp.end();++i){
if(i->second > 1)
cout<<i->first<<" ";
}
}
int[] a = {5, 6, 8, 9, 3, 4, 2, 9 };
int[] b = {5, 6, 8, 9, 3, 6, 1, 9 };
for (int i = 0; i < a.Length; i++)
{
if (a[i] != b[i])
{
Console.Write("Original Array manipulated at position {0} + "\t\n"
+ "and the element is {1} replaced by {2} ", i,
a[i],b[i] + "\t\n" );
break;
}
}
Console.Read();
///use break if want to check only one manipulation in original array.
///If want to check more then one manipulation in original array, remove break
This video If Programming Was An Anime is too fun not to share. It is the same problem and the video has the answers:
Sorting
Creating a hashmap/dictionary.
Creating an array. (Though this is partially skipped over.)
Using the Tortoise and Hare Algorithm.
Note: This problem is more of a trivia problem than it is real world. Any solution beyond a hashmap is premature optimization, except in rare limited ram situations, like embedded programming.
Furthermore, when is the last time you've seen in the real world an array where all of the variables within the array fit within the size of the array? Eg, if the data in the array is bytes (0-255) when do you have an array 256 elements or larger without nulls or inf within it, and you need to find a duplicate number? This scenario is so rare you will probably never get to use this trick in your entire career.
Because it is a trivia problem and is not real world the question, I'd be cautious accepting an offer from a company that asks trivia questions like this, because people will pass the interview by sheer luck instead of skill. This implies the devs there are not guaranteed to be skilled, which unless you're okay teaching your seniors skills, you might have a bad time.
int a[] = {2,1,2,3,4};
int b[] = {0};
for(int i = 0; i < a.size; i++)
{
if(a[i] == a[i+1])
{
//duplicate found
//copy it to second array
b[i] = a[i];
}
}
In this case, the MAX is only 5, so I could check the duplicates one by one, but how could I do this in a simpler way? For example, what if the MAX has a value of 20?
Thanks.
int MAX = 5;
for (i = 1 , i <= MAX; i++)
{
drawNum[1] = (int)(Math.random()*MAX)+1;
while (drawNum[2] == drawNum[1])
{
drawNum[2] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[3] == drawNum[1]) || (drawNum[3] == drawNum[2]) )
{
drawNum[3] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[4] == drawNum[1]) || (drawNum[4] == drawNum[2]) || (drawNum[4] == drawNum[3]) )
{
drawNum[4] = (int)(Math.random()*MAX)+1;
}
while ((drawNum[5] == drawNum[1]) ||
(drawNum[5] == drawNum[2]) ||
(drawNum[5] == drawNum[3]) ||
(drawNum[5] == drawNum[4]) )
{
drawNum[5] = (int)(Math.random()*MAX)+1;
}
}
The simplest way would be to create a list of the possible numbers (1..20 or whatever) and then shuffle them with Collections.shuffle. Then just take however many elements you want. This is great if your range is equal to the number of elements you need in the end (e.g. for shuffling a deck of cards).
That doesn't work so well if you want (say) 10 random elements in the range 1..10,000 - you'd end up doing a lot of work unnecessarily. At that point, it's probably better to keep a set of values you've generated so far, and just keep generating numbers in a loop until the next one isn't already present:
if (max < numbersNeeded)
{
throw new IllegalArgumentException("Can't ask for more numbers than are available");
}
Random rng = new Random(); // Ideally just create one instance globally
// Note: use LinkedHashSet to maintain insertion order
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
Integer next = rng.nextInt(max) + 1;
// As we're adding to a set, this will automatically do a containment check
generated.add(next);
}
Be careful with the set choice though - I've very deliberately used LinkedHashSet as it maintains insertion order, which we care about here.
Yet another option is to always make progress, by reducing the range each time and compensating for existing values. So for example, suppose you wanted 3 values in the range 0..9. On the first iteration you'd generate any number in the range 0..9 - let's say you generate a 4.
On the second iteration you'd then generate a number in the range 0..8. If the generated number is less than 4, you'd keep it as is... otherwise you add one to it. That gets you a result range of 0..9 without 4. Suppose we get 7 that way.
On the third iteration you'd generate a number in the range 0..7. If the generated number is less than 4, you'd keep it as is. If it's 4 or 5, you'd add one. If it's 6 or 7, you'd add two. That way the result range is 0..9 without 4 or 6.
Here's how I'd do it
import java.util.ArrayList;
import java.util.Random;
public class Test {
public static void main(String[] args) {
int size = 20;
ArrayList<Integer> list = new ArrayList<Integer>(size);
for(int i = 1; i <= size; i++) {
list.add(i);
}
Random rand = new Random();
while(list.size() > 0) {
int index = rand.nextInt(list.size());
System.out.println("Selected: "+list.remove(index));
}
}
}
As the esteemed Mr Skeet has pointed out:
If n is the number of randomly selected numbers you wish to choose and N is the total sample space of numbers available for selection:
If n << N, you should just store the numbers that you have picked and check a list to see if the number selected is in it.
If n ~= N, you should probably use my method, by populating a list containing the entire sample space and then removing numbers from it as you select them.
//random numbers are 0,1,2,3
ArrayList<Integer> numbers = new ArrayList<Integer>();
Random randomGenerator = new Random();
while (numbers.size() < 4) {
int random = randomGenerator .nextInt(4);
if (!numbers.contains(random)) {
numbers.add(random);
}
}
This would be a lot simpler in java-8:
Stream.generate(new Random()::ints)
.flatMap(IntStream::boxed)
.distinct()
.limit(16) // whatever limit you might need
.toArray(Integer[]::new);
There is another way of doing "random" ordered numbers with LFSR, take a look at:
http://en.wikipedia.org/wiki/Linear_feedback_shift_register
with this technique you can achieve the ordered random number by index and making sure the values are not duplicated.
But these are not TRUE random numbers because the random generation is deterministic.
But depending your case you can use this technique reducing the amount of processing on random number generation when using shuffling.
Here a LFSR algorithm in java, (I took it somewhere I don't remeber):
public final class LFSR {
private static final int M = 15;
// hard-coded for 15-bits
private static final int[] TAPS = {14, 15};
private final boolean[] bits = new boolean[M + 1];
public LFSR() {
this((int)System.currentTimeMillis());
}
public LFSR(int seed) {
for(int i = 0; i < M; i++) {
bits[i] = (((1 << i) & seed) >>> i) == 1;
}
}
/* generate a random int uniformly on the interval [-2^31 + 1, 2^31 - 1] */
public short nextShort() {
//printBits();
// calculate the integer value from the registers
short next = 0;
for(int i = 0; i < M; i++) {
next |= (bits[i] ? 1 : 0) << i;
}
// allow for zero without allowing for -2^31
if (next < 0) next++;
// calculate the last register from all the preceding
bits[M] = false;
for(int i = 0; i < TAPS.length; i++) {
bits[M] ^= bits[M - TAPS[i]];
}
// shift all the registers
for(int i = 0; i < M; i++) {
bits[i] = bits[i + 1];
}
return next;
}
/** returns random double uniformly over [0, 1) */
public double nextDouble() {
return ((nextShort() / (Integer.MAX_VALUE + 1.0)) + 1.0) / 2.0;
}
/** returns random boolean */
public boolean nextBoolean() {
return nextShort() >= 0;
}
public void printBits() {
System.out.print(bits[M] ? 1 : 0);
System.out.print(" -> ");
for(int i = M - 1; i >= 0; i--) {
System.out.print(bits[i] ? 1 : 0);
}
System.out.println();
}
public static void main(String[] args) {
LFSR rng = new LFSR();
Vector<Short> vec = new Vector<Short>();
for(int i = 0; i <= 32766; i++) {
short next = rng.nextShort();
// just testing/asserting to make
// sure the number doesn't repeat on a given list
if (vec.contains(next))
throw new RuntimeException("Index repeat: " + i);
vec.add(next);
System.out.println(next);
}
}
}
Another approach which allows you to specify how many numbers you want with size and the min and max values of the returned numbers
public static int getRandomInt(int min, int max) {
Random random = new Random();
return random.nextInt((max - min) + 1) + min;
}
public static ArrayList<Integer> getRandomNonRepeatingIntegers(int size, int min,
int max) {
ArrayList<Integer> numbers = new ArrayList<Integer>();
while (numbers.size() < size) {
int random = getRandomInt(min, max);
if (!numbers.contains(random)) {
numbers.add(random);
}
}
return numbers;
}
To use it returning 7 numbers between 0 and 25.
ArrayList<Integer> list = getRandomNonRepeatingIntegers(7, 0, 25);
for (int i = 0; i < list.size(); i++) {
System.out.println("" + list.get(i));
}
The most efficient, basic way to have non-repeating random numbers is explained by this pseudo-code. There is no need to have nested loops or hashed lookups:
// get 5 unique random numbers, possible values 0 - 19
// (assume desired number of selections < number of choices)
const int POOL_SIZE = 20;
const int VAL_COUNT = 5;
declare Array mapping[POOL_SIZE];
declare Array results[VAL_COUNT];
declare i int;
declare r int;
declare max_rand int;
// create mapping array
for (i=0; i<POOL_SIZE; i++) {
mapping[i] = i;
}
max_rand = POOL_SIZE-1; // start loop searching for maximum value (19)
for (i=0; i<VAL_COUNT; i++) {
r = Random(0, max_rand); // get random number
results[i] = mapping[r]; // grab number from map array
mapping[r] = max_rand; // place item past range at selected location
max_rand = max_rand - 1; // reduce random scope by 1
}
Suppose first iteration generated random number 3 to start (from 0 - 19). This would make results[0] = mapping[3], i.e., the value 3. We'd then assign mapping[3] to 19.
In the next iteration, the random number was 5 (from 0 - 18). This would make results[1] = mapping[5], i.e., the value 5. We'd then assign mapping[5] to 18.
Now suppose the next iteration chose 3 again (from 0 - 17). results[2] would be assigned the value of mapping[3], but now, this value is not 3, but 19.
This same protection persists for all numbers, even if you got the same number 5 times in a row. E.g., if the random number generator gave you 0 five times in a row, the results would be: [ 0, 19, 18, 17, 16 ].
You would never get the same number twice.
Generating all the indices of a sequence is generally a bad idea, as it might take a lot of time, especially if the ratio of the numbers to be chosen to MAX is low (the complexity becomes dominated by O(MAX)). This gets worse if the ratio of the numbers to be chosen to MAX approaches one, as then removing the chosen indices from the sequence of all also becomes expensive (we approach O(MAX^2/2)). But for small numbers, this generally works well and is not particularly error-prone.
Filtering the generated indices by using a collection is also a bad idea, as some time is spent in inserting the indices into the sequence, and progress is not guaranteed as the same random number can be drawn several times (but for large enough MAX it is unlikely). This could be close to complexity O(k n log^2(n)/2), ignoring the duplicates and assuming the collection uses a tree for efficient lookup (but with a significant constant cost k of allocating the tree nodes and possibly having to rebalance).
Another option is to generate the random values uniquely from the beginning, guaranteeing progress is being made. That means in the first round, a random index in [0, MAX] is generated:
items i0 i1 i2 i3 i4 i5 i6 (total 7 items)
idx 0 ^^ (index 2)
In the second round, only [0, MAX - 1] is generated (as one item was already selected):
items i0 i1 i3 i4 i5 i6 (total 6 items)
idx 1 ^^ (index 2 out of these 6, but 3 out of the original 7)
The values of the indices then need to be adjusted: if the second index falls in the second half of the sequence (after the first index), it needs to be incremented to account for the gap. We can implement this as a loop, allowing us to select arbitrary number of unique items.
For short sequences, this is quite fast O(n^2/2) algorithm:
void RandomUniqueSequence(std::vector<int> &rand_num,
const size_t n_select_num, const size_t n_item_num)
{
assert(n_select_num <= n_item_num);
rand_num.clear(); // !!
// b1: 3187.000 msec (the fastest)
// b2: 3734.000 msec
for(size_t i = 0; i < n_select_num; ++ i) {
int n = n_Rand(n_item_num - i - 1);
// get a random number
size_t n_where = i;
for(size_t j = 0; j < i; ++ j) {
if(n + j < rand_num[j]) {
n_where = j;
break;
}
}
// see where it should be inserted
rand_num.insert(rand_num.begin() + n_where, 1, n + n_where);
// insert it in the list, maintain a sorted sequence
}
// tier 1 - use comparison with offset instead of increment
}
Where n_select_num is your 5 and n_number_num is your MAX. The n_Rand(x) returns random integers in [0, x] (inclusive). This can be made a bit faster if selecting a lot of items (e.g. not 5 but 500) by using binary search to find the insertion point. To do that, we need to make sure that we meet the requirements.
We will do binary search with the comparison n + j < rand_num[j] which is the same as n < rand_num[j] - j. We need to show that rand_num[j] - j is still a sorted sequence for a sorted sequence rand_num[j]. This is fortunately easily shown, as the lowest distance between two elements of the original rand_num is one (the generated numbers are unique, so there is always difference of at least 1). At the same time, if we subtract the indices j from all the elements rand_num[j], the differences in index are exactly 1. So in the "worst" case, we get a constant sequence - but never decreasing. The binary search can therefore be used, yielding O(n log(n)) algorithm:
struct TNeedle { // in the comparison operator we need to make clear which argument is the needle and which is already in the list; we do that using the type system.
int n;
TNeedle(int _n)
:n(_n)
{}
};
class CCompareWithOffset { // custom comparison "n < rand_num[j] - j"
protected:
std::vector<int>::iterator m_p_begin_it;
public:
CCompareWithOffset(std::vector<int>::iterator p_begin_it)
:m_p_begin_it(p_begin_it)
{}
bool operator ()(const int &r_value, TNeedle n) const
{
size_t n_index = &r_value - &*m_p_begin_it;
// calculate index in the array
return r_value < n.n + n_index; // or r_value - n_index < n.n
}
bool operator ()(TNeedle n, const int &r_value) const
{
size_t n_index = &r_value - &*m_p_begin_it;
// calculate index in the array
return n.n + n_index < r_value; // or n.n < r_value - n_index
}
};
And finally:
void RandomUniqueSequence(std::vector<int> &rand_num,
const size_t n_select_num, const size_t n_item_num)
{
assert(n_select_num <= n_item_num);
rand_num.clear(); // !!
// b1: 3578.000 msec
// b2: 1703.000 msec (the fastest)
for(size_t i = 0; i < n_select_num; ++ i) {
int n = n_Rand(n_item_num - i - 1);
// get a random number
std::vector<int>::iterator p_where_it = std::upper_bound(rand_num.begin(), rand_num.end(),
TNeedle(n), CCompareWithOffset(rand_num.begin()));
// see where it should be inserted
rand_num.insert(p_where_it, 1, n + p_where_it - rand_num.begin());
// insert it in the list, maintain a sorted sequence
}
// tier 4 - use binary search
}
I have tested this on three benchmarks. First, 3 numbers were chosen out of 7 items, and a histogram of the items chosen was accumulated over 10,000 runs:
4265 4229 4351 4267 4267 4364 4257
This shows that each of the 7 items was chosen approximately the same number of times, and there is no apparent bias caused by the algorithm. All the sequences were also checked for correctness (uniqueness of contents).
The second benchmark involved choosing 7 numbers out of 5000 items. The time of several versions of the algorithm was accumulated over 10,000,000 runs. The results are denoted in comments in the code as b1. The simple version of the algorithm is slightly faster.
The third benchmark involved choosing 700 numbers out of 5000 items. The time of several versions of the algorithm was again accumulated, this time over 10,000 runs. The results are denoted in comments in the code as b2. The binary search version of the algorithm is now more than two times faster than the simple one.
The second method starts being faster for choosing more than cca 75 items on my machine (note that the complexity of either algorithm does not depend on the number of items, MAX).
It is worth mentioning that the above algorithms generate the random numbers in ascending order. But it would be simple to add another array to which the numbers would be saved in the order in which they were generated, and returning that instead (at negligible additional cost O(n)). It is not necessary to shuffle the output: that would be much slower.
Note that the sources are in C++, I don't have Java on my machine, but the concept should be clear.
EDIT:
For amusement, I have also implemented the approach that generates a list with all the indices 0 .. MAX, chooses them randomly and removes them from the list to guarantee uniqueness. Since I've chosen quite high MAX (5000), the performance is catastrophic:
// b1: 519515.000 msec
// b2: 20312.000 msec
std::vector<int> all_numbers(n_item_num);
std::iota(all_numbers.begin(), all_numbers.end(), 0);
// generate all the numbers
for(size_t i = 0; i < n_number_num; ++ i) {
assert(all_numbers.size() == n_item_num - i);
int n = n_Rand(n_item_num - i - 1);
// get a random number
rand_num.push_back(all_numbers[n]); // put it in the output list
all_numbers.erase(all_numbers.begin() + n); // erase it from the input
}
// generate random numbers
I have also implemented the approach with a set (a C++ collection), which actually comes second on benchmark b2, being only about 50% slower than the approach with the binary search. That is understandable, as the set uses a binary tree, where the insertion cost is similar to binary search. The only difference is the chance of getting duplicate items, which slows down the progress.
// b1: 20250.000 msec
// b2: 2296.000 msec
std::set<int> numbers;
while(numbers.size() < n_number_num)
numbers.insert(n_Rand(n_item_num - 1)); // might have duplicates here
// generate unique random numbers
rand_num.resize(numbers.size());
std::copy(numbers.begin(), numbers.end(), rand_num.begin());
// copy the numbers from a set to a vector
Full source code is here.
Your problem seems to reduce to choose k elements at random from a collection of n elements. The Collections.shuffle answer is thus correct, but as pointed out inefficient: its O(n).
Wikipedia: Fisher–Yates shuffle has a O(k) version when the array already exists. In your case, there is no array of elements and creating the array of elements could be very expensive, say if max were 10000000 instead of 20.
The shuffle algorithm involves initializing an array of size n where every element is equal to its index, picking k random numbers each number in a range with the max one less than the previous range, then swapping elements towards the end of the array.
You can do the same operation in O(k) time with a hashmap although I admit its kind of a pain. Note that this is only worthwhile if k is much less than n. (ie k ~ lg(n) or so), otherwise you should use the shuffle directly.
You will use your hashmap as an efficient representation of the backing array in the shuffle algorithm. Any element of the array that is equal to its index need not appear in the map. This allows you to represent an array of size n in constant time, there is no time spent initializing it.
Pick k random numbers: the first is in the range 0 to n-1, the second 0 to n-2, the third 0 to n-3 and so on, thru n-k.
Treat your random numbers as a set of swaps. The first random index swaps to the final position. The second random index swaps to the second to last position. However, instead of working against a backing array, work against your hashmap. Your hashmap will store every item that is out of position.
int getValue(i)
{
if (map.contains(i))
return map[i];
return i;
}
void setValue(i, val)
{
if (i == val)
map.remove(i);
else
map[i] = val;
}
int[] chooseK(int n, int k)
{
for (int i = 0; i < k; i++)
{
int randomIndex = nextRandom(0, n - i); //(n - i is exclusive)
int desiredIndex = n-i-1;
int valAtRandom = getValue(randomIndex);
int valAtDesired = getValue(desiredIndex);
setValue(desiredIndex, valAtRandom);
setValue(randomIndex, valAtDesired);
}
int[] output = new int[k];
for (int i = 0; i < k; i++)
{
output[i] = (getValue(n-i-1));
}
return output;
}
You could use one of the classes implementing the Set interface (API), and then each number you generate, use Set.add() to insert it.
If the return value is false, you know the number has already been generated before.
Instead of doing all this create a LinkedHashSet object and random numbers to it by Math.random() function .... if any duplicated entry occurs the LinkedHashSet object won't add that number to its List ... Since in this Collection Class no duplicate values are allowed .. in the end u get a list of random numbers having no duplicated values .... :D
With Java 8 upwards you can use the ints method from the IntStream interface:
Returns an effectively unlimited stream of pseudorandom int values.
Random r = new Random();
int randomNumberOrigin = 0;
int randomNumberBound = 10;
int size = 5;
int[] unique = r.ints(randomNumberOrigin, randomNumberBound)
.distinct()
.limit(size)
.toArray();
Following code create a sequence random number between [1,m] that was not generated before.
public class NewClass {
public List<Integer> keys = new ArrayList<Integer>();
public int rand(int m) {
int n = (int) (Math.random() * m + 1);
if (!keys.contains(n)) {
keys.add(n);
return n;
} else {
return rand(m);
}
}
public static void main(String[] args) {
int m = 4;
NewClass ne = new NewClass();
for (int i = 0; i < 4; i++) {
System.out.println(ne.rand(m));
}
System.out.println("list: " + ne.keys);
}
}
The most easy way is use nano DateTime as long format.
System.nanoTime();
There is algorithm of card batch: you create ordered array of numbers (the "card batch") and in every iteration you select a number at random position from it (removing the selected number from the "card batch" of course).
Here is an efficient solution for fast creation of a randomized array. After randomization you can simply pick the n-th element e of the array, increment n and return e. This solution has O(1) for getting a random number and O(n) for initialization, but as a tradeoff requires a good amount of memory if n gets large enough.
There is a more efficient and less cumbersome solution for integers than a Collections.shuffle.
The problem is the same as successively picking items from only the un-picked items in a set and setting them in order somewhere else. This is exactly like randomly dealing cards or drawing winning raffle tickets from a hat or bin.
This algorithm works for loading any array and achieving a random order at the end of the load. It also works for adding into a List collection (or any other indexed collection) and achieving a random sequence in the collection at the end of the adds.
It can be done with a single array, created once, or a numerically ordered collectio, such as a List, in place. For an array, the initial array size needs to be the exact size to contain all the intended values. If you don't know how many values might occur in advance, using a numerically orderred collection, such as an ArrayList or List, where the size is not immutable, will also work. It will work universally for an array of any size up to Integer.MAX_VALUE which is just over 2,000,000,000. List objects will have the same index limits. Your machine may run out of memory before you get to an array of that size. It may be more efficient to load an array typed to the object types and convert it to some collection, after loading the array. This is especially true if the target collection is not numerically indexed.
This algorithm, exactly as written, will create a very even distribution where there are no duplicates. One aspect that is VERY IMPORTANT is that it has to be possible for the insertion of the next item to occur up to the current size + 1. Thus, for the second item, it could be possible to store it in location 0 or location 1. For the 20th item, it could be possible to store it in any location, 0 through 19. It is just as possible the first item to stay in location 0 as it is for it to end up in any other location. It is just as possible for the next new item to go anywhere, including the next new location.
The randomness of the sequence will be as random as the randomness of the random number generator.
This algorithm can also be used to load reference types into random locations in an array. Since this works with an array, it can also work with collections. That means you don't have to create the collection and then shuffle it or have it ordered on whatever orders the objects being inserted. The collection need only have the ability to insert an item anywhere in the collection or append it.
// RandomSequence.java
import java.util.Random;
public class RandomSequence {
public static void main(String[] args) {
// create an array of the size and type for which
// you want a random sequence
int[] randomSequence = new int[20];
Random randomNumbers = new Random();
for (int i = 0; i < randomSequence.length; i++ ) {
if (i == 0) { // seed first entry in array with item 0
randomSequence[i] = 0;
} else { // for all other items...
// choose a random pointer to the segment of the
// array already containing items
int pointer = randomNumbers.nextInt(i + 1);
randomSequence[i] = randomSequence[pointer];
randomSequence[pointer] = i;
// note that if pointer & i are equal
// the new value will just go into location i and possibly stay there
// this is VERY IMPORTANT to ensure the sequence is really random
// and not biased
} // end if...else
} // end for
for (int number: randomSequence) {
System.out.printf("%2d ", number);
} // end for
} // end main
} // end class RandomSequence
It really all depends on exactly WHAT you need the random generation for, but here's my take.
First, create a standalone method for generating the random number.
Be sure to allow for limits.
public static int newRandom(int limit){
return generatedRandom.nextInt(limit); }
Next, you will want to create a very simple decision structure that compares values. This can be done in one of two ways. If you have a very limited amount of numbers to verify, a simple IF statement will suffice:
public static int testDuplicates(int int1, int int2, int int3, int int4, int int5){
boolean loopFlag = true;
while(loopFlag == true){
if(int1 == int2 || int1 == int3 || int1 == int4 || int1 == int5 || int1 == 0){
int1 = newRandom(75);
loopFlag = true; }
else{
loopFlag = false; }}
return int1; }
The above compares int1 to int2 through int5, as well as making sure that there are no zeroes in the randoms.
With these two methods in place, we can do the following:
num1 = newRandom(limit1);
num2 = newRandom(limit1);
num3 = newRandom(limit1);
num4 = newRandom(limit1);
num5 = newRandom(limit1);
Followed By:
num1 = testDuplicates(num1, num2, num3, num4, num5);
num2 = testDuplicates(num2, num1, num3, num4, num5);
num3 = testDuplicates(num3, num1, num2, num4, num5);
num4 = testDuplicates(num4, num1, num2, num3, num5);
num5 = testDuplicates(num5, num1, num2, num3, num5);
If you have a longer list to verify, then a more complex method will yield better results both in clarity of code and in processing resources.
Hope this helps. This site has helped me so much, I felt obliged to at least TRY to help as well.
I created a snippet that generates no duplicate random integer. the advantage of this snippet is that you can assign the list of an array to it and generate the random item, too.
No duplication random generator class
With Java 8 using the below code, you can create 10 distinct random Integer Numbers within a range of 1000.
Random random = new Random();
Integer[] input9 = IntStream.range(1, 10).map(i -> random.nextInt(1000)).boxed().distinct()
.toArray(Integer[]::new);
System.out.println(Arrays.toString(input9));
Modify the range to generate more numbers example : range(1,X). It will generate X distinct random numbers.
Modify the nextInt value to select the random number range : random.nextInt(Y)::random number will be generated within the range Y