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Leetcode 454. 4Sum II

Problem Statement

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

Clarifying Questions

  1. What is the length of each input array?
    • All input arrays are of the same length n.
  2. Are there any constraints on the values within the arrays?
    • The values can be positive, negative, or zero and there may be no specific constraints other than standard integer limits.
  3. Do we need to handle any specific edge cases?
    • Empty arrays are not specified, so we can assume n is at least 1.
    • Since the problem deals with potentially large integers due to summing, confirm that we don’t need to handle overflow explicitly in Java.

Strategy

We can use a HashMap to store the sums of all possible pairs from two of the arrays (nums1 and nums2), and then for each possible pair sum from the remaining two arrays (nums3 and nums4), we check if the negation of that sum exists in the HashMap.

  1. Step 1: Create a HashMap to store the sum of pairs of elements from nums1 and nums2.
  2. Step 2: Iterate through nums3 and nums4, calculate the sum of each pair and check if the negation of that sum exists in the HashMap. If it exists, it means there is a corresponding pair in nums1 and nums2 that can make the overall sum zero.
  3. Step 3: Sum up the counts and return the total number of such tuples.

Code

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer, Integer> sumCountMap = new HashMap<>();
        
        // Calculate all pair sums from nums1 and nums2, and store them in the map
        for (int num1 : nums1) {
            for (int num2 : nums2) {
                int sum = num1 + num2;
                sumCountMap.put(sum, sumCountMap.getOrDefault(sum, 0) + 1);
            }
        }
        
        int count = 0;
        
        // Calculate all pair sums from nums3 and nums4, and check against the map
        for (int num3 : nums3) {
            for (int num4 : nums4) {
                int sum = num3 + num4;
                count += sumCountMap.getOrDefault(-sum, 0);
            }
        }
        
        return count;
    }
}

Time Complexity

The time complexity of the algorithm is as follows:

Thus, the overall time complexity is O(n^2) + O(n^2) = O(n^2).

Space Complexity

The space complexity is O(n^2) for storing the pair sums in the HashMap.

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