Given a number and an array find all possible combinations that sum up to given number

Question

I was asked about this program in an interview. The task is given an integer array as input and another number say target , find all possible combinations in array that sum up to given target.

Example:

array = [1,2,1,1,1]
target = 3

expected result : [1,2],[1,1,1],[2,1],[2,1]

I have come up with below code:

public static void main(String args[]) {
    Integer[] numbers = { 1, 2, 1, 1, 1 };
    int target = 3;
    process(numbers, target);
}

private static void process(Integer[] numbers, int target) {
    for (int i = 0; i < numbers.length; i++) {
        int sum = 0;

        for (int j = i; j < numbers.length; j++) {
            sum += numbers[j];
            if (sum == target) {
                for (int k = i; k <= j; k++) {
                    System.out.print(numbers[k] + " ");
                }
                System.out.println();
            }
        }
    }
}

But this code only prints 3 combinations : [1 2], [2 1], [1 1 1]

What is the correct way to do this? Is there any other better solution.

Answer 1

You can do something like this using backtracking:

  public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> temp = new ArrayList<>();
    Arrays.sort(candidates);
    int start = 0; 

    backtrack(candidates, target, start, result, temp);

    return result;
}

public void backtrack(int[] candidates, int remaining, int start, List<List<Integer>> result, List<Integer> temp) {
    if(remaining == 0) { 
        result.add(new ArrayList<>(temp)); 
        return;
    }

    if(remaining < 0) { return; }

    for(int i = start; i < candidates.length; i++) {
        temp.add(candidates[i]);
        backtrack(candidates, remaining - candidates[i], i, result, temp);
        temp.remove(temp.size() - 1);
    }
}

You can read more about backtracking and similar approaches.

Answer 2

This is an excellent candidate for using recursion:

public static void main(String args[]) {
    Integer[] numbers = { 1, 2, 1, 1, 1 };
    int target = 3;

    for(int i = 0; i < numbers.length; i++) {
        recurseAdd(i, numbers, new ArrayList<Integer>(), 0, target);
    }
}

private static void recurseAdd(int currentIndex, Integer[] numbers, List<Integer> usedNumbers, int sum, int target) {
    if (currentIndex >= numbers.length) {
        return;
    }

    sum = sum + numbers[currentIndex];
    usedNumbers.add(numbers[currentIndex]);

    if (sum == target) {
        System.out.println(usedNumbers);
        return;
    }

    if (sum > target) {
        return;
    }

   for (int i = currentIndex + 1; i < numbers.length; i++) {
        recurseAdd(i, numbers, new ArrayList<>(usedNumbers), sum, target);
    }
}