Question

454. 4Sum II

中文文档

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: HashMap

Time complexity $O(n^2)$, Space complexity $O(n^2)$.

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)

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;
  }
}

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;
  }
};

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
}

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