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 < nnums1[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.lengthn == nums2.lengthn == nums3.lengthn == nums4.length1 <= 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.
PythonJavaC++GoTypeScriptRust
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)(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
