454. 4Sum II

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

思路Coding化:

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        
        count = {}
        #兩兩一組，先建立相加後有幾個
        for i in nums1:
            for j in nums2:
                count[i+j] = count.get(i+j,0) + 1
        #再去看剩下的nums3 nums4中，有哪些總和等於 -(i+j)
        total = 0
        for i in nums3:
            for j in nums4:
                total += count.get(-(i+j), 0)
        return total

完整代碼:

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        
        count = {}
        for i in nums1:
            for j in nums2:
                count[i+j] = count.get(i+j,0) + 1
        
        total = 0
        for i in nums3:
            for j in nums4:
                total += count.get(-(i+j), 0)
        return total

Time Complexity: O(n²)

Space Complexity: O(n²)