How can I find two elements in an array that sum to k

Question

Suppose you are given an array of unsorted integers as

A = {3,4,5,1,4,2}

Input : 6 Output : {5,1}, {4,2}

Answer 1

My little answer about this problem for O(n) complexity with O(n) additional memory. This code snippet returns all unique indices pair of elements with sum K.

/**
 * Returns indices of all complementary pairs in given {@code arr} with factor {@code k}. Two elements {@code arr[i]} and {@code arr[j]} are
 * complementary if {@code arr[i] + arr[j] = k}.
 * Method returns set of pairs in format {@literal [i,j]}. Two pairs {@literal [i,j]} and {@literal [j,i]} are treated as same, and only one pair
 * is returned.
 * Method support negative numbers in the {@code arr}, as wel as negative {@code k}.
 * 

 * Complexity of this method is O(n), requires O(n) additional memory.
 *
 * @param arr source array
 * @param k   factor number
 * @return not {@code null} set of all complementary pairs in format {@literal [i,j]}
 */
public static Set getComplementaryPairs(int[] arr, int k) {
    if (arr == null || arr.length == 0)
        return Collections.emptySet();

    Map> indices = new TreeMap<>();

    for (int i = 0; i < arr.length; i++) {
        if (!indices.containsKey(arr[i]))
            indices.put(arr[i], new TreeSet<>());
        indices.get(arr[i]).add(i);
    }

    Set res = new LinkedHashSet<>();

    for (Map.Entry> entry : indices.entrySet()) {
        int x = entry.getKey();
        int y = k - x;

        if (x == y) {
            int size = entry.getValue().size();

            if (size < 2)
                continue;

            Integer[] ind = entry.getValue().toArray(new Integer[size]);

            for (int j = 0; j < size - 1; j++)
                for (int m = j + 1; m < size; m++)
                    res.add(String.format("[%d,%d]", ind[j], ind[m]));
        } else if (x < y && indices.containsKey(y))
            for (int j : entry.getValue())
                for (int m : indices.get(y))
                    res.add(String.format("[%d,%d]", j, m));
    }

    return res;
}