Leetcode 4Sum II problem solution YASH PAL, 31 July 2024 In this Leetcode 4Sum II problem solution we have given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that: 0 <= i, j, k, l < n nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 Problem solution in Python. from itertools import product class Solution: def fourSumCount(self, A: List[int], B: List[int], C: List[int], D: List[int]) -> int: def merge(x, y): rest = {} for i in product(x, y): m = sum(i) if m in rest: rest[m] += 1 else: rest[m] = 1 return rest r1 = merge(A, B) r2 = merge(C, D) rest = 0 for i1, i2 in product(r1.keys(), r2.keys()): m = i1 + i2 if m == 0: rest += r1[i1] * r2[i2] return rest Problem solution in Java. class Solution { public int fourSumCount(int[] A, int[] B, int[] C, int[] D) { int count=0; HashMap<Integer,Integer> map=new HashMap<>(); for(int i=0;i<A.length;i++){ for(int j=0;j<B.length;j++){ int sum=A[i]+B[j]; map.put(sum, map.getOrDefault(sum,0)+1); } } for(int i=0;i<C.length;i++){ for(int j=0;j<D.length;j++){ int sum=C[i]+D[j]; if(map.containsKey(-1*sum)){ count+=map.get(-1*sum); } } } return count; } } Problem solution in C++. class Solution { public: int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) { int count=0; unordered_map<int,int>mp; for(int i=0;i<nums1.size();i++){ for(int j=0;j<nums2.size();j++){ mp[nums1[i]+nums2[j]]++; } } for(int k=0;k<nums3.size();k++){ for(int l=0;l<nums4.size();l++){ count+=mp[-(nums3[k]+nums4[l])]; } } return count; } }; coding problems