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Tested Python solution for LeetCode 454 with 16 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 454 # by problem number
lcpy gen -s four_sum_ii # by problem name
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] == 0Input: 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
Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1
n == nums1.lengthn == nums2.lengthn == nums3.lengthn == nums4.length1 <= n <= 200-2^28 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2^28class Solution:
# Time: O(n^2)
# Space: O(n^2)
def four_sum_count(
self, nums1: list[int], nums2: list[int], nums3: list[int], nums4: list[int]
) -> int:
pair_sums: dict[int, int] = {}
for a in nums1:
for b in nums2:
pair_sums[a + b] = pair_sums.get(a + b, 0) + 1
count = 0
for c in nums3:
for d in nums4:
count += pair_sums.get(-(c + d), 0)
return count
| Time | Space |
|---|---|
| O(n^2) | O(n^2) |