Develop Reference

Verifying the time complexity of selection sort by counting the number of comparisons

I'm trying to verify that selection sort is O(n^2) by counting the number of times numbers are being compared in my array. I'm using an array of 20 numbers so I know I should be getting somewhere around 400, but with the current placement of my counting variable I'm only getting around 200 comparisons being counted. Here is the code for the selection sort I am using:
public static void selectionSort(final int[] arr) {
for (int i = 0; i < arr.length - 1; i++) {
int minElementIndex = i;
for (int j = i + 1; j < arr.length; j++) {
swapCount++;
if (arr[minElementIndex] > arr[j])
{
minElementIndex = j;
}
}
if (minElementIndex != i) {
int temp = arr[i];
arr[i] = arr[minElementIndex];
arr[minElementIndex] = temp;
}
}
}
The swapCount variable is being used to track the comparisons and I've tried moving it around everywhere I could think of, but I can't get a count nearly as high as I'm expecting when I print the total swapCount at the end. Where exactly in the code should swapCount be to track all the comparisons?

Largest number in k swaps

Given a number K and string str of digits denoting a positive integer, build the largest number possible by performing swap operations on the digits of str at most K times.
Example 1:
Input:
K = 4
str = "1234567"
Output:
7654321
Explanation:
Three swaps can make the
input 1234567 to 7654321, swapping 1
with 7, 2 with 6 and finally 3 with 5
I am trying to solve it using two loops. For every index i, I am finding the largest integer between (i+1)th index to (N-1)th index, where N is the size of string. If the largest number is greater than arr[i], then swap it. Below is the code I have written.
public static String findMaximumNum(String str, int k) {
int N = str.length();
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = Integer.valueOf(str.charAt(i) + "");
}
int swaps = 0;
for (int i = 0; i < N - 1; i++) {
if(swaps == k)
break;
int maxIndex = findMaxInRange(arr, i + 1, N - 1);
if (arr[i] < arr[maxIndex]) {
swap(arr, i, maxIndex);
swaps++;
}
}
String out = "";
for (int i = 0; i < N; i++) {
out = out + arr[i] + "";
}
return out;
}
private static int findMaxInRange(int[] arr, int i, int j) {
int max = Integer.MIN_VALUE;
int maxIndex = i;
for (int k = i; k <= j; k++) {
if (arr[k] >= max) {
max = arr[k];
maxIndex = k;
}
}
return maxIndex;
}
private static void swap(int[] arr, int i, int j) {
System.out.println("swapping "+arr[i]+" and "+arr[j]+" from "+Arrays.toString(arr));
int ch = arr[i];
arr[i] = arr[j];
arr[j] = ch;
}
public static void main(String[] args) {
System.out.println(findMaximumNum("61892795431", 4));
}
It is failing for few test cases. One of the test cases where it is failing is
Input:
4
61892795431
Its Correct output is:
99876215431
And MyCode's output is:
99876125431
I am not able to figure out how the output is '99876215431' and what is wrong in my approach. Please help me understand. Thanks a lot in advance :)

The basic steps how to solve this problem:
0. cast string to array of integers
make a loop K times
in this loop go from i+1 (LOOP VAR) to end of a collection and search for higher value
when we find higher value then collection[i], we will remember its value and index on witch it is. Important thing to note is that we want to swap biggest number but i also has to be last possible number.
at the end of iteration we swap the elements (i with best index)
we are done so all its left is convert our int list back to string.
code: (its python because java is pain)
def sort(swaps, string):
l = list(map(int, list(string)))
print(l)
for i in range(swaps):
best = l[i] + 1
bestIndex = i
for j in range(i+1, len(l)):
if best <= l[j]:
best = l[j]
bestIndex = j
print(i, bestIndex)
l[i], l[bestIndex] = l[bestIndex], l[i]
return "".join(map(str, l))
print(sort(4, "61892795431"))

Your code is correct. The problem comes from the parameter 4 (max number of swaps). If you use 10, the sorting is completed successfully.
Maybe the deeper problems comes from the fact that you are comparing the swaps of your algorithm with the swaps that you would do efficiently to sort the numbers. Your algorithm may work but probably it is not the most efficient, so the number of swaps needed is above the optimum.

What is the complexity of these two string based algorithms?

I have written these two algorithms to check a string for duplicate characters (ABBC, AAAC). The first uses the hashset data structure, whilst the second relies purely on iteration.
Algorithm 1
String s = "abcdefghijklmnopqrstuvwxxyz";
public boolean isUnique(String s) {
Set<Character> charSet = new HashSet<Character>();
for(int i=0; i<s.length(); i++) {
if(charSet.contains(s.charAt(i))) {
return false;
}
charSet.add(s.charAt(i));
}
return true;
}
Algorithm 2
String s = "abcdefghijklmnopqrstuvwxxyz";
public boolean isUnique2(String s) {
for(int i=0; i<s.length()-1; i++) {
for(int j = i+1; j<s.length(); j++) {
if(s.charAt(i) == s.charAt(j)) {
return false;
}
}
}
return true;
}
My thoughts are that the first algorithm is O(N), and the second algorithm is O(N^2). When i run execution time tests on my (possibly unreliable) laptop, the average speed for the first algorithm is 2020ns, whilst the second algorithm is 995ns. This goes against my calculation of the algorithms complexity, could anybody advise me?

when using O() notation you ignore constants, which means that O(n) == (10^10*n). so while O(n^2)>O(n) is true asymptotically, its not necessarily true for smaller values of n.
in your case imagine that maybe resizing the array behind the hashset could be more time consuming than iterating the input.

Assuming that the charAt method runs in O(1) time, the first algorithm is O(N) and the second is O(N^2). A linear time algorithm is not supposed to be faster than a quadratic algorithm for all inputs. It will be faster than the quadratic one after a certain N (which could possibly be in the millions).
for example:
void funcA(int n){
for (int i = 0; i < n; i++){
for (int j = 0; j < 10000; j++){
int k = i + j;
}
}
}
void funcB(int n){
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
int k = i + j;
}
}
}
even though funcA is linear and funcB is quadratic, it is easy to see that funcB will be faster than funcA for n < 10000. In your case the hashSet requires time to compute the hash and so may be slower for inputs of a certain size.

Micro benchmarking that you are doing can give very misleading info about algorithm complexities.
It's easy to "port" your algorithms to check for duplicates in, say, array of Integers.
Then I recommend testing performance on say, array of 10^7 elements and you will definitely see the difference.
This way you'd be able to confirm your initially correct estimation O(N) for hashset vs O(N^2) for the second "loop" version.

There is a problem with your test data, for example, if you limit yourself to English language characters (a-z), you are guaranteed to have a duplicate if string length > 26. In the specific example you provided the string "abcdefghijklmnopqrstuvwxxyz" is sorted and the duplicate element x is found towards the end. As such iterated array lookup is faster because there is overhead in building the HashSet as you continue to parse the String.
A better test would be to test this with randomly generated integer sequences of large sizes and with a large max value, e.g. Long.MAX_VALUE
Below is a test that disproves your assertion that array search is faster. Run it a few times and see for yourself. Or you could take averages from 1000 runs, etc:
public class FindDuplicatesTest {
public static final String s = generateRandomString(100000);
private static String generateRandomString(int numChars) {
Random random = new Random();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < numChars; i++) {
int codePoint = random.nextInt(65536);
sb.append(Character.toChars(codePoint));
}
return sb.toString();
}
public boolean isUnique(String s) {
Set<Character> charSet = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
if (charSet.contains(s.charAt(i))) {
return false;
}
charSet.add(s.charAt(i));
}
return true;
}
public boolean isUnique2(String s) {
for (int i = 0; i < s.length() - 1; i++) {
for (int j = i + 1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
return false;
}
}
}
return true;
}
public static void main(String[] args) {
FindDuplicatesTest app = new FindDuplicatesTest();
long start = System.nanoTime();
boolean result = app.isUnique(s);
long stop = System.nanoTime();
System.out.println(result);
System.out.println("HashSet Search Time: " + (stop - start));
start = System.nanoTime();
result = app.isUnique2(s);
stop = System.nanoTime();
System.out.println(result);
System.out.println("Array Search Time: " + (stop - start));
}
}

Get indices of n maximums in java array

I have an array of size 1000. How can I find the indices (indexes) of the five maximum elements?
An example with setup code and my attempt are displayed below:
Random rand = new Random();
int[] myArray = new int[1000];
int[] maxIndices = new int[5];
int[] maxValues = new int[5];
for (int i = 0; i < myArray.length; i++) {
myArray[i] = rand.nextInt();
}
for (int i = 0; i < 5; i++) {
maxIndices[i] = i;
maxValues[i] = myArray[i];
}
for (int i = 0; i < maxIndices.length; i++) {
for (int j = 0; j < myArray.length; j++) {
if (myArray[j] > maxValues[i]) {
maxIndices[i] = j;
maxValues[i] = myArray[j];
}
}
}
for (int i = 0; i < maxIndices.length; i++) {
System.out.println("Index: " + maxIndices[i]);
}
I know the problem is that it is constantly assigning the highest maximum value to all the maximum elements. I am unsure how to remedy this because I have to preserve the values and the indices of myArray.
I don't think sorting is an option because I need to preserve the indices. In fact, it is the indices that I need specifically.

Sorry to answer this old question but I am missing an implementation which has all following properties:
Easy to read
Performant
Handling of multiple same values
Therefore I implemented it:
private int[] getBestKIndices(float[] array, int num) {
//create sort able array with index and value pair
IndexValuePair[] pairs = new IndexValuePair[array.length];
for (int i = 0; i < array.length; i++) {
pairs[i] = new IndexValuePair(i, array[i]);
}
//sort
Arrays.sort(pairs, new Comparator<IndexValuePair>() {
public int compare(IndexValuePair o1, IndexValuePair o2) {
return Float.compare(o2.value, o1.value);
}
});
//extract the indices
int[] result = new int[num];
for (int i = 0; i < num; i++) {
result[i] = pairs[i].index;
}
return result;
}
private class IndexValuePair {
private int index;
private float value;
public IndexValuePair(int index, float value) {
this.index = index;
this.value = value;
}
}

Sorting is an option, at the expense of extra memory. Consider the following algorithm.
1. Allocate additional array and copy into - O(n)
2. Sort additional array - O(n lg n)
3. Lop off the top k elements (in this case 5) - O(n), since k could be up to n
4. Iterate over the original array - O(n)
4.a search the top k elements for to see if they contain the current element - O(lg n)
So it step 4 is (n * lg n), just like the sort. The entire algorithm is n lg n, and is very simple to code.
Here's a quick and dirty example. There may be bugs in it, and obviously null checking and the like come into play.
import java.util.Arrays;
class ArrayTest {
public static void main(String[] args) {
int[] arr = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10};
int[] indexes = indexesOfTopElements(arr,3);
for(int i = 0; i < indexes.length; i++) {
int index = indexes[i];
System.out.println(index + " " + arr[index]);
}
}
static int[] indexesOfTopElements(int[] orig, int nummax) {
int[] copy = Arrays.copyOf(orig,orig.length);
Arrays.sort(copy);
int[] honey = Arrays.copyOfRange(copy,copy.length - nummax, copy.length);
int[] result = new int[nummax];
int resultPos = 0;
for(int i = 0; i < orig.length; i++) {
int onTrial = orig[i];
int index = Arrays.binarySearch(honey,onTrial);
if(index < 0) continue;
result[resultPos++] = i;
}
return result;
}
}
There are other things you can do to reduce the overhead of this operation. For example instead of sorting, you could opt to use a queue that just tracks the largest 5. Being ints they values would probably have to be boxed to be added to a collection (unless you rolled your own) which adds to overhead significantly.

a bit late in answering, you could also use this function that I wrote:
/**
* Return the indexes correspond to the top-k largest in an array.
*/
public static int[] maxKIndex(double[] array, int top_k) {
double[] max = new double[top_k];
int[] maxIndex = new int[top_k];
Arrays.fill(max, Double.NEGATIVE_INFINITY);
Arrays.fill(maxIndex, -1);
top: for(int i = 0; i < array.length; i++) {
for(int j = 0; j < top_k; j++) {
if(array[i] > max[j]) {
for(int x = top_k - 1; x > j; x--) {
maxIndex[x] = maxIndex[x-1]; max[x] = max[x-1];
}
maxIndex[j] = i; max[j] = array[i];
continue top;
}
}
}
return maxIndex;
}

My quick and a bit "think outside the box" idea would be to use the EvictingQueue that holds an maximum of 5 elements. You'd had to pre-fill it with the first five elements from your array (do it in a ascending order, so the first element you add is the lowest from the five).
Than you have to iterate through the array and add a new element to the queue whenever the current value is greater than the lowest value in the queue. To remember the indexes, create a wrapper object (a value/index pair).
After iterating through the whole array, you have your five maximum value/index pairs in the queue (in descending order).
It's a O(n) solution.

Arrays.sort(myArray), then take the final 5 elements.
Sort a copy if you want to preserve the original order.
If you want the indices, there isn't a quick-and-dirty solution as there would be in python or some other languages. You sort and scan, but that's ugly.
Or you could go objecty - this is java, after all.
Make an ArrayMaxFilter object. It'll have a private class ArrayElement, which consists of an index and a value and has a natural ordering by value. It'll have a method which takes a pair of ints, index and value, creates an ArrayElement of them, and drops them into a priority queue of length 5. (or however many you want to find). Submit each index/value pair from the array, then report out the values remaining in the queue.
(yes, a priority queue traditionally keeps the lowest values, but you can flip this in your implementation)

Here is my solution. Create a class that pairs an indice with a value:
public class IndiceValuePair{
private int indice;
private int value;
public IndiceValuePair(int ind, int val){
indice = ind;
value = val;
}
public int getIndice(){
return indice;
}
public int getValue(){
return value;
}
}
and then use this class in your main method:
public static void main(String[] args){
Random rand = new Random();
int[] myArray = new int[10];
IndiceValuePair[] pairs = new IndiceValuePair[5];
System.out.println("Here are the indices and their values:");
for(int i = 0; i < myArray.length; i++) {
myArray[i] = rand.nextInt(100);
System.out.println(i+ ": " + myArray[i]);
for(int j = 0; j < pairs.length; j++){
//for the first five entries
if(pairs[j] == null){
pairs[j] = new IndiceValuePair(i, myArray[i]);
break;
}
else if(pairs[j].getValue() < myArray[i]){
//inserts the new pair into its correct spot
for(int k = 4; k > j; k--){
pairs[k] = pairs [k-1];
}
pairs[j] = new IndiceValuePair(i, myArray[i]);
break;
}
}
}
System.out.println("\n5 Max indices and their values");
for(int i = 0; i < pairs.length; i++){
System.out.println(pairs[i].getIndice() + ": " + pairs[i].getValue());
}
}
and example output from a run:
Here are the indices and their values:
0: 13
1: 71
2: 45
3: 38
4: 43
5: 9
6: 4
7: 5
8: 59
9: 60
5 Max indices and their values
1: 71
9: 60
8: 59
2: 45
4: 43
The example I provided only generates ten ints with a value between 0 and 99 just so that I could see that it worked. You can easily change this to fit 1000 values of any size. Also, rather than run 3 separate for loops, I checked to see if the newest value I add is a max value right after I add to to myArray. Give it a run and see if it works for you

Create an int array of int arrays that contains all possible sums that add up to a given number

I'm completely new in Java. I am writing an Android game, and I need to generate an array of int arrays that contains all possible sums (excluding combinations that contains number 2 or is bigger than 8 numbers) that add up to a given number.
For example:
ganeratePatterns(5) must return array
[patternNumber][summandNumber] = value
[0][0] = 5
[1][0] = 1
[1][1] = 1
[1][2] = 1
[1][3] = 1
[1][4] = 1
[2][0] = 3
[2][1] = 1
[2][2] = 1
[3][0] = 4
[3][1] = 1
I already try to do this like there Getting all possible sums that add up to a given number
but it's very difficult to me to make it like this http://introcs.cs.princeton.edu/java/23recursion/Partition.java.html
Solution
int n = 10;
int dimension = 0;
//First we need to count number of posible combinations to create a 2dimensionarray
for(List<Integer> sumt : new SumIterator(n)) {
if(!sumt.contains(2) && sumt.size() < 9) {
dimension++;
}
}
int[][] combinationPattern = new int[dimension][];
int foo = 0;
for(List<Integer> sum : new SumIterator(n)) {
if(!sum.contains(2) && sum.size() < 9) {
System.out.println(sum);
combinationPattern[foo] = toIntArray(sum);
foo++;
}
}
It's work not 100% correctly, and very pretty, but it is enough for my game
I have used SumIterator class from here SumIterator.class
I have to changed this code for(int j = n-1; j > n/2; j--) { to this for(int j = n-1; j >= n/2; j--) { because old version doesn't return all combinations (like [5,5] for 10)
And I used toIntArray function. I have founded hare on StackOverflow, but forget a link so here it's source:
public static int[] toIntArray(final Collection<Integer> data){
int[] result;
// null result for null input
if(data == null){
result = null;
// empty array for empty collection
} else if(data.isEmpty()){
result = new int[0];
} else{
final Collection<Integer> effective;
// if data contains null make defensive copy
// and remove null values
if(data.contains(null)){
effective = new ArrayList<Integer>(data);
while(effective.remove(null)){}
// otherwise use original collection
}else{
effective = data;
}
result = new int[effective.size()];
int offset = 0;
// store values
for(final Integer i : effective){
result[offset++] = i.intValue();
}
}
return result;
}

This is not the most beautiful code, but it does what you would like, having modified the code you referenced. It is also quite fast. It could be made faster by staying away from recursion (using a stack), and completely avoiding String-to-integer conversion. I may come back and edit those changes in. Running on my pretty outdated laptop, it printed the partitions of 50 (all 204226 of them) in under 5 seconds.
When partition(N) exits in this code, partitions will hold the partitions of N.
First, it builds an ArrayList of string representations of the sums in space-delimited format (example: " 1 1 1").
It then creates a two-dimensional array of ints which can hold all of the results.
It splits each String in the ArrayList into an array of Strings which each contain only a single number.
For each String, it creates an array of ints by parsing each number into an array.
This int array is then added to the two-dimensional array of ints.
Let me know if you have any questions!
import java.util.ArrayList;
public class Partition
{
static ArrayList<String> list = new ArrayList<String>();
static int[][] partitions;
public static void partition(int n)
{
partition(n, n, "");
partitions = new int[list.size()][0];
for (int i = 0; i < list.size(); i++)
{
String s = list.get(i);
String[] stringAsArray = s.trim().split(" ");
int[] intArray = new int[stringAsArray.length];
for (int j = 0; j < stringAsArray.length; j++)
{
intArray[j] = Integer.parseInt(stringAsArray[j]);
}
partitions[i] = intArray;
}
}
public static void partition(int n, int max, String prefix)
{
if(prefix.trim().split(" ").length > 8 || (prefix + " ").contains(" 2 "))
{
return;
}
if (n == 0)
{
list.add(prefix);
return;
}
for (int i = Math.min(max, n); i >= 1; i--)
{
partition(n - i, i, prefix + " " + i);
}
}
public static void main(String[] args)
{
int N = 50;
partition(N);
/**
* Demonstrates that the above code works as intended.
*/
for (int i = 0; i < partitions.length; i++)
{
int[] currentArray = partitions[i];
for (int j = 0; j < currentArray.length; j++)
{
System.out.print(currentArray[j] + " ");
}
System.out.println();
}
}
}