maximum_subarr_sum_len_div_k.py (870B)
| # 3381. Maximum Subarray Sum With Length Divisible by K
from itertools import accumulate
"""
you are given an array of integers 'nums' and an integer 'k'. return the
maximum sum of a subarray of 'nums', such that the size of the subarray is
divisble by 'k'.
"""
class Solution(object):
def maxSubarraySum(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
pre, ans = k * [float("inf")], float("-inf")
pre[-1] = 0
for i, j in enumerate(accumulate(nums)):
ans = max(ans, j - pre[i % k])
pre[i % k] = min(pre[i % k], j)
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.maxSubarraySum(nums=[1, 2], k=1))
print(obj.maxSubarraySum(nums=[-1, -2, -3, -4, -5], k=4))
print(obj.maxSubarraySum(nums=[-5, 1, 2, -3, 4], k=2))
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