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Home/ Questions/Q 8175929
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T22:57:02+00:00 2026-06-06T22:57:02+00:00

I’ve been looking around quite a bit to solve my issue. I got many

  • 0

I’ve been looking around quite a bit to solve my issue. I got many problems solved but this one is still haunting me :S It’s been a long time I haven’t touch Java programming (programming in general) so be understanding out there! 😉

My goal is to get all the combination possible out of an array of integers. When I use the following code, applied to the test array of integer {1, 2, 3, 4}, I expect to have:

1 2 3 4

1 2 4 3

1 3 2 4

1 3 4 2

2 1 3 4

2 1 4 3

(…)

but here is what I get

1 2 3 4

1 2 3 4 4 3

1 2 3 4 4 3 3 2 4

Does anybody have a clue, a suggestion or even a solution? Thanks in advance!

public class Calculation{
(...)
  public void Permute(ArrayList<Integer> soFar,ArrayList<Integer> rest){
    if(rest.isEmpty())    this.fillMatrice(convertIntegers(soFar)); // there it goes in a previously created arrow of int
    else{
        for(int k=0;k<rest.size();k++){
            ArrayList<Integer> next=new ArrayList<Integer>();
            next=soFar;
            next.add(rest.get(k));
            ArrayList<Integer> remaining=new ArrayList<Integer>();
            List<Integer> sublist = rest.subList(0, k);
            for(int a=0;a<sublist.size();a++)   remaining.add(sublist.get(a));
            sublist = rest.subList(k+1,rest.size());
            for(int a=0;a<sublist.size();a++)   remaining.add(sublist.get(a));
            Permute(next,remaining);
        }
    }
}
public static ArrayList<Integer> convertArray(int[] integers){
    ArrayList<Integer> convArray=new ArrayList<Integer>();
    for(int i=0;i<integers.length;i++)  convArray.add(integers[i]);
    return convArray;
}
public static int[] convertIntegers(List<Integer> integers){
    int[] ret = new int[integers.size()];
    for(int i=0;i<ret.length;i++)   ret[i]=integers.get(i).intValue();
    return ret;
}
public Calculation() {
    (...)
    ArrayList<Integer> soFar=new ArrayList<Integer>();
    int[] test={1,2,3,4};
    Permute(soFar,convertArray(test));
}
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  1. Editorial Team
    Editorial Team
    2026-06-06T22:57:04+00:00Added an answer on June 6, 2026 at 10:57 pm

    You can try Recursion to solve this issue:

    public static void printPermutations(int[] n, int[] Nr, int idx) {
        if (idx == n.length) {  //stop condition for the recursion [base clause]
            System.out.println(Arrays.toString(n));
            return;
        }
        for (int i = 0; i <= Nr[idx]; i++) { 
            n[idx] = i;
            printPermutations(n, Nr, idx+1); //recursive invokation, for next elements
        }
    }
    

    More info can be had from this link:
    Combinatorics: generate all “states” – array combinations

    You can replicate the same logic here as well.

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