454. 4Sum Ii¶

Difficulty: Medium

454. 4Sum II

Medium

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

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1

Constraints:

n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
-2²⁸ <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2²⁸

Solution¶

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        int n = nums1.length, count = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int ele1 : nums3) {
            for (int ele2 : nums4) {
                map.put((ele1 + ele2) , map.getOrDefault(ele1 + ele2 , 0) + 1);
            } 
        }
        for (int ele1 : nums1) {
            for (int ele2 : nums2) {
                int req = 0 - (ele1 + ele2);
                count += map.getOrDefault(req, 0);
            }
        }
        return count;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here