0454 - 4Sum II (Medium)

Problem Link

https://leetcode.com/problems/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

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
• -2^28 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2^28

Approach 1: Brute Force (TLE)

The simple way is to iterate over all $nums$ to construct all possible tuples. However, this solution gives TLE as it is a $O(n^4)$ solution.

Written by @wingkwong

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int ans = 0;
        for (auto a : nums1) {
            for (auto b : nums2) {
                for (auto c : nums3) {
                    for (auto d : nums4) {
                        ans += a + b + c + d == 0;
                    }
                }
            }
        }
        return ans;
    }
};

Approach 2: Hasp Map (TLE)

We can take the same approach from 0001 - Two Sum (Easy). We know that $a + b + c + d = 0$ is same as $a = -(b + c + d)$. Therefore, we iterate $nums1$ to count the number of element $a$. Then iterate $nums2$, $nums3$ and $nums4$ to find if the complementary sum $-(b + c + d)$ is in hash map or not. However, this solution gives TLE as it is a $O(n^3)$ solution.

Written by @wingkwong

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int ans = 0;
        unordered_map<int, int> m;
        for (auto a : nums1) m[a]++;
        for (auto b : nums2) {
            for (auto c : nums3) {
                for (auto d : nums4) {
                    // a + b + c + d == 0
                    // a = -(b + c + d)
                    ans += m[-(b + c + d)];
                }
            }
        }
        return ans;
    }
};

Approach 3: Hash Map (Accepted)

We can take the same approach from 0001 - Two Sum (Easy). We know that $a + b + c + d = 0$ is same as $a + b = - (c + d)$. Therefore, we iterate $nums1$and $nums2$ to count the sum of elements $a + b$. Then iterate $nums3$ and $nums4$to find if the complementary sum $- (c + d)$is in hash map or not. This solution passes as it is a $O(n^2)$solution.

Written by @wingkwong

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int ans = 0;
        unordered_map<int, int> m;
        for (auto a : nums1) {
            for (auto b : nums2) {
                m[a + b]++;
            }
        }
        for (auto c : nums3) {
            for (auto d : nums4) {
                // a + b + c + d == 0
                // a + b = -(c + d)
                ans += m[-(c + d)];
            }
        }
        return ans;
    }
};

You may also use two hash maps to store the sums $a + b$ and $-(c + d)$ and multiply the values at the end.

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        int ans = 0;
        unordered_map<int, int> m1, m2;
        for (auto a : nums1) {
            for (auto b : nums2) {
                m1[a + b]++;
            }
        }
        for (auto c : nums3) {
            for (auto d : nums4) {
                m2[-(c + d)]++;
            }
        }
        for (auto x : m1) {
            ans += m1[x.first] * m2[x.first];
        }
        return ans;
    }
};