/*
* @lc app=leetcode id=454 lang=javascript
*
* [454] 4Sum II
*
* https://leetcode.com/problems/4sum-ii/description/
/**
* @param {number[]} A
* @param {number[]} B
* @param {number[]} C
* @param {number[]} D
* @return {number}
*/
var fourSumCount = function (A, B, C, D) {
const sumMapper = {};
let res = 0;
for (let i = 0; i < A.length; i++) {
for (let j = 0; j < B.length; j++) {
sumMapper[A[i] + B[j]] = (sumMapper[A[i] + B[j]] || 0) + 1;
}
}
for (let i = 0; i < C.length; i++) {
for (let j = 0; j < D.length; j++) {
res += sumMapper[-(C[i] + D[j])] || 0;
}
}
return res;
};
Python3:
class Solution:
def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int:
mapper = {}
res = 0
for i in A:
for j in B:
mapper[i + j] = mapper.get(i + j, 0) + 1
for i in C:
for j in D:
res += mapper.get(-1 * (i + j), 0)
return res
