# 689. Maximum Sum of 3 Non-Overlapping Subarrays
"""
given an integer array 'nums' and an integer 'k', find three non-overlapping
subarrays of length 'k' with maximum sum and return them. return the result
as a list of indices representing the starting position of each interval
(0-indexed). if there are multiple answers, return the lexicographcially
smallest one
"""
class Solution(object):
def max_sumOfThreeSubarrays(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
n = len(nums)
max_sum = 0
sum = [0]
left = [0] * n
right = [n - k] * n
ans = [0, 0, 0]
for i in nums:
sum.append(sum[-1] + i)
tot = sum[k] - sum[0]
for i in range(k, n):
if sum[i + 1] - sum[i + 1 - k] > tot:
left[i] = i + 1 - k
tot = sum[i + 1] - sum[i + 1 - k]
else:
left[i] = left[i - 1]
tot = sum[n] - sum[n - k]
for i in range(n - k - 1, -1, -1):
if sum[i + k] - sum[i] > tot:
right[i] = i
tot = sum[i + k] - sum[i]
else:
right[i] = right[i + 1]
for i in range(k, n - 2 * k + 1):
l, r = left[i - 1], right[i + k]
tot = (sum[i + k] - sum[i]) + (sum[l + k] - sum[l]) + (sum[r + k] - sum[r])
if tot > max_sum:
max_sum = tot
ans = [l, i, r]
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.max_sumOfThreeSubarrays(nums=[1, 2, 1, 2, 6, 7, 5, 1], k=2))
print(obj.max_sumOfThreeSubarrays(nums=[1, 2, 1, 2, 1, 2, 1, 2, 1], k=2))
