Sponsored
Sponsored
This approach leverages hash maps to efficiently count and use pair sums. First, we compute all possible sums between arrays nums1 and nums2, storing them in a hash map along with their count. Then, we iterate over pairs of nums3 and nums4, checking how many complements (to form a sum of zero) exist in the hash map.
Time Complexity: O(n^2), with each pair calculation taking constant time.
Space Complexity: O(n^2), using space for hash map storage.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int FourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
6 Dictionary<int, int> sumMap = new Dictionary<int, int>();
7 int count = 0;
8
9 foreach (int a in nums1) {
10 foreach (int b in nums2) {
11 int sum = a + b;
12 if (sumMap.ContainsKey(sum)) {
13 sumMap[sum]++;
14 } else {
15 sumMap[sum] = 1;
16 }
17 }
18 }
19
20 foreach (int c in nums3) {
21 foreach (int d in nums4) {
22 int complement = -(c + d);
23 if (sumMap.ContainsKey(complement)) {
24 count += sumMap[complement];
25 }
26 }
27 }
28
29 return count;
30 }
31}
32This C# solution uses a Dictionary to store the sums of pairs from the first two arrays. As it computes pairs from nums3 and nums4, it sums any complementary values found in the dictionary to obtain total valid tuples.