Sub set sum solution java

Question

I need to write a code using recursion and backtracking and without any loops that will find all possible solutions for the equation x1+x2+x3 = K where K is a given number. and x1 , x2, x3 are non zero positive integers between 1 - 10 .

My attempt:

public static int subSetSum(int i, int k, int A[]) {
    int sum = A[0] + A[1] + A[2];
    int noOfSolutions = 0;
    if(k - sum < 0 || i >= A.length)
        return 0;
    if(k - sum == 0) {
        System.out.println(A[0] + " + " + A[1] + " + " + A[2]);
        noOfSolutions =+ 1; }
    noOfSolutions = subSetSum(i+1,k,A);
    int newA[] = A;
    newA[i] = A[i]+1;
    noOfSolutions = subSetSum(i,k,newA);
    return noOfSolutions;
}

Running the code I will only get the one solution. So If try to find all solutions for 6 it will only print out 1+1+4 and 0 (for no' of solutions).

Answer 1

• (1) If you want to add and assign the operator is +=, not =+ (this will assign +1 to noOfSolutions)
• (2) You don't need a new vector and also it will be the same (A and newA it will have the same memory address, so a change for one of them will affect both of them)
• (3) You should revert your stack changes after the recursive call

Edit

• (4) You should stop after a solution

public static int subSetSum(int i, int k, int A[]) {
    int sum = A[0] + A[1] + A[2];
    int noOfSolutions = 0;
    if(k - sum < 0 || i >= A.length)
        return 0;
    if(k - sum == 0) {
        System.out.println(A[0] + " + " + A[1] + " + " + A[2]);
--(1)-->    noOfSolutions += 1;
--(4)-->    return noOfSolutions;
    }
    noOfSolutions += subSetSum(i+1,k,A);
--(2)-->    A[i] = A[i]+1;
    noOfSolutions += subSetSum(i,k,A);
--(3)-->    A[i] = A[i]-1;
    return noOfSolutions;
}

Exemple

public static void main(String[] args) {
    System.out.println(subSetSum(0, 4, new int[3]));
}

Output

0 + 0 + 4
0 + 1 + 3
0 + 2 + 2
0 + 3 + 1
0 + 4 + 0
1 + 0 + 3
1 + 1 + 2
1 + 2 + 1
1 + 3 + 0
2 + 0 + 2
2 + 1 + 1
2 + 2 + 0
3 + 0 + 1
3 + 1 + 0
4 + 0 + 0
15