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:
- (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
- (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))
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