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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.
1function fourSumCount(nums1, nums2, nums3, nums4) {
2 const sumMap = new Map();
3 let count = 0;
4
5 for (let a of nums1) {
6 for (let b of nums2) {
7 const sum = a + b;
8 sumMap.set(sum, (sumMap.get(sum) || 0) + 1);
9 }
10 }
11
12 for (let c of nums3) {
13 for (let d of nums4) {
14 const complement = -(c + d);
15 if (sumMap.has(complement)) {
16 count += sumMap.get(complement);
17 }
18 }
19 }
20
21 return count;
22}
23The JavaScript implementation employs a Map to keep counts of the sums of elements from nums1 and nums2. It then iterates through possible sums of nums3 and nums4, checking how many can complement each other to form a zero total sum.