Algorithms for Finding all Possible Combinations of k Elements in an Array with Java Implementation

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java-servlet-json Given an array of size N e.g. e={'A','B','C','D','E'} N=5, we want to find all possible combinations of k elements in that array. For example, if k=3 then one possible combination is {'A','B','C'}. Here we have three different algorithms for finding k-combinations of an array.

Forward-Backward Algorithm

Here we have two arrays and two main indices r & i:

  • Array e which is the elements array.
  • Array pointers which is an array for holding indices for selected element.
  • Index i for pointing to current selected element in array e.
  • Index r for pointing to current position in pointers array.
  • The algorithm will move forward by incrementing i & r as long as they do not exceed arrays length.
  • If r reaches the last position of pointers array a combination is printed.
  • If both indices reach the last poisition of their pointing arrays the algorith will step backward by reducing r value r-- and set i with the value of i = pointer[r]+1.


public static void combination(Object[]  elements, int k){

	// get the length of the array
	// e.g. for {'A','B','C','D'} => N = 4 
	int N = elements.length;
	if(k > N){
		System.out.println("Invalid input, K > N");
	// init combination index array
	int pointers[] = new int[k];

	int r = 0; // index for combination array
	int i = 0; // index for elements array
	while(r >= 0){
		// forward step if i < (N + (r-K))
		if(i <= (N + (r - k))){
			pointers[r] = i;
			// if combination array is full print and increment i;
			if(r == k-1){
				print(pointers, elements);
				// if combination is not full yet, select next element
				i = pointers[r]+1;
		// backward step
			if(r >= 0)
				i = pointers[r]+1;

Shifting Algorithm

  • This algorithm is more intuitive than the first one.
  • We virtually split the elements array into two types of elements, k elements that can be selected and N-k elements that will be ignored.
  • In each iteration we select N-k non-ignored elements.
  • After each iteration we shift the positions of ignored elements as shown in the image below.


public static void combination(Object[]  e, int k){

	int[] ignore = new int[e.length-k]; // --> [0][0]
	int[] combination = new int[k]; // --> [][][]
	// set initial ignored elements 
	//(last k elements will be ignored)
	for(int w = 0; w < ignore.length; w++)
		ignore[w] = e.length - k +(w+1);
	int i = 0, r = 0, g = 0;
	boolean terminate = false;
		// selecting N-k non-ignored elements
		while(i < e.length && r < k){
    		if(i != ignore[g]){
    			combination[r] = i;
    			r++; i++;
    			if(g != ignore.length-1)
    	print(combination, e);
    	i = 0; r = 0; g = 0;

    	terminate = true;
    	// shifting ignored indices
    	for(int w = 0 ; w < ignore.length; w++){
    		if(ignore[w] > w){	    			
    			if(w > 0)
    				ignore[w-1] = ignore[w]-1;
    			terminate = false;

Recursive Algorithm

  • Recursive algorithm has shorter steps.
  • In each call to the function we pass List of elements, k and an accumulated combination.
  • Then we have four conditions:
    1. if elements.length < k then stop
    2. if k == 1 then add each element to the accumulated combination
    3. if elements.length == k then add all elements to the accumulated combination.
    4. if elements.length > k then for each element e make a recursive call passing sub list of the elements list, k-1 and add element e to accumulated combination.
  • It works as shown below


public static void combination(List<String> e, int k, String accumulated){

	// 1. stop
	if(e.size() < k)
	// 2. add each element in e to accumulated
	if(k == 1)
		for(String s:e)
	// 3. add all elements in e to accumulated
	else if(e.size() == k){
		for(String s:e)
	// 4. for each element, call combination
	else if(e.size() > k)
		for(int i = 0 ; i < e.size() ; i++)
			combination(e.subList(i+1, e.size()), k-1, accumulated+e.get(i));

Source Code @ GitHub

Exploring Kotlin

Less than 100 pages covering Kotlin syntax and features in straight and to the point explanation.