4Sum II

LeetCode Q 454 - 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 1.

Solution

The solution is quite similar to Two Sum.
We use a HashMap to store the possible sum of every pair in A and B.
Then we check if the sum of some pair in C and D equals to -key in the map.

Time Complexity: O(n^2)

Code:

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
	<Integer, Integer> map = new HashMap<>();
	for (int a: A) {
		for (int b: B) {
			map.put(a + b, map.getOrDefault(a + b, 0) + 1);
		}
	}

	int res = 0;
	for (int c: C) {
		for (int d: D) {
			if (map.containsKey(-c-d)) res += map.get(-c-d);
		}
	}

	return res;
}