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Abhishek Chaudhary

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# 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

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`

SOLUTION:

``````class Solution:
def numTuples(self, k, i, n):
if (i, k) in self.cache:
return self.cache[(i, k)]
if i == n - 1:
ctr = self.nums[i].count(k)
self.cache[(i, k)] = ctr
return ctr
else:
ctr = 0
for num in self.nums[i]:
ctr += self.numTuples(k - num, i + 1, n)
self.cache[(i, k)] = ctr
return ctr

def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
self.cache = {}
self.nums = [nums1, nums2, nums3, nums4]
return self.numTuples(0, 0, len(self.nums))
``````