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];
}
}
I've been having some trouble with this code I made up, in that I've been getting an ArrayIndexOutofBounds exception error and I have no idea what is going on. What I want my code to produce is a char array of numbers.
Here's the code:
public class testing {
static int k = 2;
public static void main(String args[]) {
int[] numArr = new int[] { 15, 18, 21 };
createChar(numArr, k);
}
public static char[] createChar(int numArr[], int k) {
char[] store = new char[k - 1];
for (int i = 0; i < k; i++) {
int divide = numArr[i] / 4;
int numStore = divide % 4;
charN(numStore, store, k);
}
return store;
}
public static char[] charN(int numStore, char store[], int k) {
for (int i = 0; i < k; i++) {
if (k == 0) {
} else {
store[k - 1] = (char) numStore;
k = k - 1;
}
}
System.out.print(store);
return store;
}
}
Thanks a lot!
As the error suggests, it is due to access outside the bounds of the array. ie. some bad index access on an array.
char[] store = new char[k - 1];
You are creating a store character array of size k-1
and passing k as size to charN()
charN(numStore, store, k);
To resolve your problem.
change store declaration as below
char[] store = new char[k];
Also, in class Testing, k must be 3, not 2. where k is the size of the array.
static int k = 3;
When size of the array is k, the indices of array run from 0 to k-1
For value k = 2, the below statement: -
char[] store = new char[k - 1];
creates an array of size 1. So, you can only access its 0th index.
But, further in the charN method, you are accessing it's 1st index at: -
store[k - 1] = (char) numStore; // Since k = 2, k - 1 = 1.
Change your array creation to: -
char[] store = new char[k];
Furthermore, I don't understand why you took k = 2 on first place. May be you meant it to be used as index, in which case, your array creation would be of size k + 1. And accordingly, your for loop will iterate till k + 1, rather than k.
Also, I don't understand the role of your charN method. In fact, tt seems strange that first you are iterating over your array in createChar method, and then passing each element to charN method. And then there also you are iterating over the char array, assigning the same value to multiple index. Apart from that, you are decrementing k and incrementing i at the same time in the loop. That is really strange. And at the end, you are returning your array from both of your methods, but are not using the return value at all.
It's quite hard to understand what you want to do. But you should consider all the points I stated in the previous paragraph. For each step there, think of why you want to do this? Is there any alternative, which might be easy? Do, I really need to methods with 2 iteration here?
I suggest you to take a look at your design once again. The solution which I posted above might solve your compiler error, but there is some problem with your logic. You need to take care of that.
As per your post the statement
char[] store = new char[k - 1];
results in a char array of size 1.
So in charN method when you try to access the
store[k - 1] = (char) numStore;
you are trying to access store[1] as 'k' is 2 here. Which is wrong. because with store[1] you are trying to access the second element in the array where as the array size is only 1.
Here's what the layout is
index num
0 [10]
1 [20]
2 [30]
(Add 35 here)
3 [40] Move elements down
4 [50]
5 [60]
6 [70]
then my method is this
public static void method(int[] num, int index, int addnum)
{
}
How can i add 35 in there?
Tried this:
public static void method(int[] num, int index, int addnum)
{
int index = 10;
for(int k = num.length k>3; k++)
{
Num[k]=num[k++]
}
Num[3] = 35;
As this is something you should accomplish yourself, I will only provide the method to implement it, not the code:
If you would set the number at position index, you would overwrite the value that was there previously. So what you need to do is move every element one position towards the end of the array starting from index: num[x] becomes num[x+1], etc.
You will find out that you need to do this in reverse order, otherwise you will fill your array with the value in num[index].
During this process you will need to decide what to do with the last entry of the array (num[num.length - 1]):
You could just overwrite it, discarding the value
You could return it from your function
You could throw an exception if it is non-zero
You could create a new array that is 1 entry larger than the current array instead to keep all values
etc.
After this, you have duplicated num[index]: the value is present in num[index+1], too, as you have moved it away.
Now it is possible to write the new value at the desired position without overriding an existing value.
EDIT
You have several errors in your code:
You increment k, you need to decrement it (k--, not k++)
You modify k again in your loop body: it is updated twice in each cycle
If you start with k = num.length, you will try to write at num[num.length + 1], which is not possible
Very crudely, you want to do something like this:
public static void(int[] num, int index, int addnum)
{
// initialize new array with size of current array plus room for new element
int[] newArray = new int[num.length + 1];
// loop until we reach point of insertion of new element
// copy the value from the same position in old array over to
// same position in new array
for(int i = 0; i < index; i++)
{
newArray[i] = num[i];
}
i = i + 1; // move to position to insert new value
newArray[i] = addnum; // insert the value
// loop until you reach the length of the old array
while(i < num.length)
{
newArray[i] = num[i-1];
}
// finally copy last value over
newArray[i + 1] = num[i];
}
You need to
allocate a new array with room for one new element.
int[] newArray = new int[oldArray.length + 1];
Copy over all elements and leave room for the one to insert.
for (int i = 0; i < newArray.length - 1; i++)
newArray[i < insertIndex ? i : i + 1] = oldArray[i];
Insert 35 in the empty spot.
newArray[insertIndex] = numberToInsert;
Note that it's not possible to do in a method like this:
public static void method(int[] num, int index, int addnum)
^^^^
since you can't change the length of num.
You need to allocate a new array, which means that need to return the new array:
public static int[] method(int[] num, int index, int addnum)
^^^^^
and then call the method like this:
myArr = method(myArr, 3, 35);
Since this very closely resembles homework what you need to realize is that you cannot dynamically increase the size of an array. So in your function:
public static void(int[] num, int index, int addnum)
{
int[] temp = new int[num.length *2];
for(int i = 0; i < index; i++)
copy num[i] into temp[i]
insert addnum into temp[index]
fill temp with remaining num values
}
That pseudocode above should get you started.
What you're looking for is an insertion sort.
It's classwork, so it's up to you to figure out the proper code.
Well, you can't unless there is "extra space" in your array, and then you can shift all elements [starting from index] one element to the right, and add 35 [num] to the relevant place.
[what actually happen is that the last element is discarded out].
However - a better solution will probably be to use an ArrayList, and use the method myArrayList.add(index,element)
How about this?
public class test {
public static void main(String[] arg) throws IOException
{
int[] myarray={1,2,3,5,6};//4 is missing we are going to add 4
int[] temp_myarray=myarray;//take a temp array
myarray=addElement(myarray,0);//increase length of myarray and add any value(I take 0) to the end
for(int i=0;i<myarray.length;i++)
{ if(i==3) //becaues I want to add the value 4 in 4th place
myarray[i]=4;
else if(i>3)
myarray[i]=temp_myarray[i-1];
else
myarray[i]=temp_myarray[i];
}
for(int i=0;i<myarray.length;i++)
System.out.print(myarray[i]);//Print new array
}
static int[] addElement(int[] arr, int elem) {
arr = Arrays.copyOf(arr, arr.length + 1);
arr[arr.length - 1] = elem;
return arr;
}
}