454. 4Sum II 

Description

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

Solutions

Solution 1: Hash Table

We can add the elements $a$ and $b$ in arrays $nums1$ and $nums2$ respectively, and store all possible sums in a hash table $cnt$, where the key is the sum of the two numbers, and the value is the count of the sum.

Then we iterate through the elements $c$ and $d$ in arrays $nums3$ and $nums4$, let $c+d$ be the target value, then the answer is the cumulative sum of $cnt[-(c+d)]$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array.

Python3

class Solution:
    def fourSumCount(
        self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]
    ) -> int:
        cnt = Counter(a + b for a in nums1 for b in nums2)
        return sum(cnt[-(c + d)] for c in nums3 for d in nums4)

Java

class Solution {
    public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int a : nums1) {
            for (int b : nums2) {
                cnt.merge(a + b, 1, Integer::sum);
            }
        }
        int ans = 0;
        for (int c : nums3) {
            for (int d : nums4) {
                ans += cnt.getOrDefault(-(c + d), 0);
            }
        }
        return ans;
    }
}

C++

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

Go

func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) (ans int) {
	cnt := map[int]int{}
	for _, a := range nums1 {
		for _, b := range nums2 {
			cnt[a+b]++
		}
	}
	for _, c := range nums3 {
		for _, d := range nums4 {
			ans += cnt[-(c + d)]
		}
	}
	return
}

TypeScript

function fourSumCount(nums1: number[], nums2: number[], nums3: number[], nums4: number[]): number {
    const cnt: Record<number, number> = {};
    for (const a of nums1) {
        for (const b of nums2) {
            const x = a + b;
            cnt[x] = (cnt[x] || 0) + 1;
        }
    }
    let ans = 0;
    for (const c of nums3) {
        for (const d of nums4) {
            const x = c + d;
            ans += cnt[-x] || 0;
        }
    }
    return ans;
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn four_sum_count(
        nums1: Vec<i32>,
        nums2: Vec<i32>,
        nums3: Vec<i32>,
        nums4: Vec<i32>,
    ) -> i32 {
        let mut cnt = HashMap::new();
        for &a in &nums1 {
            for &b in &nums2 {
                *cnt.entry(a + b).or_insert(0) += 1;
            }
        }
        let mut ans = 0;
        for &c in &nums3 {
            for &d in &nums4 {
                if let Some(&count) = cnt.get(&(0 - (c + d))) {
                    ans += count;
                }
            }
        }
        ans
    }
}