Given a sum of two elements in an int array, how to find their smallest indexes?

Question

I want to return an array of 2 integers, containing the indexes of a pair of integers in the array whose sum is k.

int[] numbers = new int[]{1, 5, 8, 1, 2};
int k = 13;

I have done this method :

public int[] findSumPair(int[] numbers, int k){
   
    int[] pairs = new int[2];

    for(int i = 0; i < numbers.length; i++){
        for(int j = i + 1; j < numbers.length; j++){
            if (numbers[i] + numbers[j] == k){
                pairs = new int[]{i, j};

               if (numbers[i] + numbers[j] != k){
                   pairs = new int[]{0, 0};
               }
            }
        }
    }
    return pairs;
}

With previous array and sum it works & returns:

[1,2]

For example with this:

int[] numbers = new int[]{1, 5, 0, 1, 2, 7, 8, 6};
int k = 13;

I should have:

[1,6] // But I get [5,7]

How can I, in case there are several possible pairs whose sum is equal to the target, return the pair with the lowest left index?

And in the case of 2 pairs with the same left index, go for the pair with the lowest right index?

Many thanks in advance

Answer 1

Assume a return of [n,m] is an valid output where n<>m and m > 0. Assume a return of [0,0] is an exception output indicates the pair is not found.

There are some mistake in your function:

    public int[] findSumPair(int[] numbers, int k) {

        int[] pairs = new int[2];

        for (int i = 0; i < numbers.length; i++) {
            for (int j = i + 1; j < numbers.length; j++) {
                if (numbers[i] + numbers[j] == k) {
                    // if a pair is found, you could stop looking anymore
                    pairs = new int[] { i, j };

                    // unreachable statement
                    if (numbers[i] + numbers[j] != k) {
                        pairs = new int[] { 0, 0 };
                    }
                }
            }
        }
        return pairs;
    }

Firstly, if you want to find the smallest index, once you found a matched pair, you could break the for-loop. Since you start searching from the left, the first match must be the smallest. No matter now hard the program try, the next result is not the smallest.

Secondly, I think you try to set a false case condition to return [0,0] if no pair is found. However, the condition is contradict to its outer if condition. That is a unreachable statement. You can put that at the very end of the function since when we reach that part, the searching is already completed. I modified the function as:

    public int[] findSumPair(int[] numbers, int k) {

        for (int i = 0; i < numbers.length; i++) {
            for (int j = i + 1; j < numbers.length; j++) {
                if (numbers[i] + numbers[j] == k) {
                    // quit immediately
                    return new int[] { i, j };
                }
            }
        }
        // if we reach here, no pair is found
        return new int[2];
    }

the main()

        int[] result = main.findSumPair(new int[] { 1, 5, 8, 1, 2 }, 13);
        System.out.println(Arrays.toString(result));

        result = main.findSumPair(new int[] { 1, 5, 0, 1, 2, 7, 8, 6 }, 13);
        System.out.println(Arrays.toString(result));

        result = main.findSumPair(new int[] { 1, 2, 2, 2, 2, 1 }, 2);
        System.out.println(Arrays.toString(result));

        result = main.findSumPair(new int[] { 1, 2, 2, 2, 4, 3 }, 7);
        System.out.println(Arrays.toString(result));

        result = main.findSumPair(new int[] { 1, 2, 3, 9, 9, 9, 4 }, 5);
        System.out.println(Arrays.toString(result));

console output

[1, 2]
[5, 7]
[0, 5]
[4, 5]
[0, 6]