# 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))