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
- code
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
d = defaultdict(int)
for n1 in nums1:
for n2 in nums2:
d[n1 + n2] += 1
res = 0
for n3 in nums3:
for n4 in nums4:
res += d[-(n3 + n4)]
return res
- code
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
d = defaultdict(int)
for n1, n2 in product(nums1, nums2):
d[n1 + n2] += 1
res = 0
for n3, n4 in product(nums3, nums4):
res += d[-(n3 + n4)]
return res
- code TLE
class Solution:
def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
def twoSum(target, n1, n2):
d = dict(Counter(n2))
count = 0
for v in n1:
if target - v in d:
count += d[target - v]
return count
def threeSum(target, n1, n2, n3):
cache = {}
ans = 0
for v in n1:
if target - v not in cache:
cache[target - v] = twoSum(target - v, n2, n3)
ans += cache[target - v]
return ans
res = 0
for v in nums1:
cache = {}
if -v not in cache:
cache[-v] = threeSum(-v, nums2, nums3, nums4)
res += cache[-v]
return res