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.

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

• nums1.length = nums2.length = nums3.length = nums4.length = n.
• 1 <= n <= 200.
• -2^28 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2^28.

Solution 1: Four loops

Code

#include <iostream>
#include <vector>
using namespace std;
int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
    int count = 0;
    for (auto& n1 : nums1) {
        for (auto& n2 : nums2) {
            for (auto& n3 : nums3) {
                for (auto& n4 : nums4) {
                    if (n1 + n2 + n3 + n4 == 0) {
                        count++;
                    }
                }
            }
        }
    }
    return count;    
}
int main() {
    vector<int> nums1, nums2, nums3, nums4;
    nums1 = {1, 2};
    nums2 = {-2, -1};
    nums3 = {-1, 2};
    nums4 = {0, 2};
    cout << fourSumCount(nums1, nums2, nums3, nums4) << endl;
    nums1 = {0};
    nums2 = {0};
    nums3 = {0};
    nums4 = {0};
    cout << fourSumCount(nums1, nums2, nums3, nums4) << endl;
}

Output:
2
1

Complexity

• Runtime: O(n^4).
• Extra memory: O(1).

Solution 2: Two loops

Rewriting nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 as nums1[i] + nums2[j] == 0 - nums3[k] - nums4[l] you could reduce the runtime complexity of the solution above from O(n^4) to O(n^2) by storing the sum nums1[i] + nums2[j] in a map.

Code

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
    int count = 0;
    unordered_map<int, int> sumTwo; 
    for (auto& n1 : nums1) {
        for (auto& n2 : nums2) {
            sumTwo[n1 + n2]++;
        }
    }
    for (auto& n3 : nums3) {
        for (auto& n4 : nums4) {
            auto it = sumTwo.find(0 - n3 - n4);
            if (it != sumTwo.end()) {
                count += it->second;
            }
        }
    }
    return count;    
}
int main() {
    vector<int> nums1, nums2, nums3, nums4;
    nums1 = {1, 2};
    nums2 = {-2, -1};
    nums3 = {-1, 2};
    nums4 = {0, 2};
    cout << fourSumCount(nums1, nums2, nums3, nums4) << endl;
    nums1 = {0};
    nums2 = {0};
    nums3 = {0};
    nums4 = {0};
    cout << fourSumCount(nums1, nums2, nums3, nums4) << endl;
}

Output:
2
1

Complexity

• Runtime: O(n^2).
• Extra memory: O(n^2).