|
| 1 | +""" |
| 2 | +Problem Link: https://leetcode.com/problems/4sum-ii/ |
| 3 | + |
| 4 | +Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples |
| 5 | +(i, j, k, l) such that: |
| 6 | +0 <= i, j, k, l < n |
| 7 | +nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0 |
| 8 | + |
| 9 | +Example 1: |
| 10 | +Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2] |
| 11 | +Output: 2 |
| 12 | +Explanation: |
| 13 | +The two tuples are: |
| 14 | +1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0 |
| 15 | +2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0 |
| 16 | + |
| 17 | +Example 2: |
| 18 | +Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0] |
| 19 | +Output: 1 |
| 20 | + |
| 21 | +Constraints: |
| 22 | +n == nums1.length |
| 23 | +n == nums2.length |
| 24 | +n == nums3.length |
| 25 | +n == nums4.length |
| 26 | +1 <= n <= 200 |
| 27 | +-228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228 |
| 28 | +""" |
| 29 | +# O(N^2) |
| 30 | +class Solution: |
| 31 | + def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int: |
| 32 | + cur_sum = {} |
| 33 | + |
| 34 | + for num1 in nums1: |
| 35 | + for num2 in nums2: |
| 36 | + cur_sum[num1+num2] = cur_sum.get(num1+num2, 0) + 1 |
| 37 | + |
| 38 | + count = 0 |
| 39 | + for num3 in nums3: |
| 40 | + for num4 in nums4: |
| 41 | + count += cur_sum.get(-num3-num4, 0) |
| 42 | + |
| 43 | + return count |
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