454. 4Sum II
Tags: ‘Hash Table’, ‘Binary Search’
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 -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
Example:
Input: A = [ 1, 2] B = [-2,-1] C = [-1, 2] D = [ 0, 2] Output: 2 Explanation: The two tuples are: 1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
Solution
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
Map<Integer, Integer> map = new HashMap<>(); // Map<Sum of AB, frequency>
for (int a : A) {
for (int b : B) {
int s = a + b;
map.put(s, map.getOrDefault(s, 0) + 1);
}
}
int result = 0;
for (int c : C) {
for (int d : D) {
if (map.containsKey(0-c-d)) {
result += map.get(0-c-d);
}
}
}
return result;
}
}