LeetCode/DynamicProgramming/MaximumSumOf3NonOverlappingSubarrays.py at master · chenjun0210/LeetCode · GitHub

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# -*- coding: utf-8 -*-
# @File    : MaximumSumOf3NonOverlappingSubarrays.py
# @Date    : 2020-08-31
# @Author  : tc
"""
题号 689. 三个无重叠子数组的最大和
给定数组 nums 由正整数组成，找到三个互不重叠的子数组的最大和。

每个子数组的长度为k，我们要使这3*k个项的和最大化。

返回每个区间起始索引的列表（索引从 0 开始）。如果有多个结果，返回字典序最小的一个。

示例:

输入: [1,2,1,2,6,7,5,1], 2
输出: [0, 3, 5]
解释: 子数组 [1, 2], [2, 6], [7, 5] 对应的起始索引为 [0, 3, 5]。
我们也可以取 [2, 1], 但是结果 [1, 3, 5] 在字典序上更大。
注意:

nums.length的范围在[1, 20000]之间。
nums[i]的范围在[1, 65535]之间。
k的范围在[1, floor(nums.length / 3)]之间。

参考：https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/discuss/108238/Python-o(n)-time-o(1)-space.-Greedy-solution.
"""
from typing import List

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        # Best single, double, and triple sequence found so far
        bestSeq = 0
        bestTwoSeq = [0, k]
        bestThreeSeq = [0, k, k * 2]

        # Sums of each window
        seqSum = sum(nums[0:k])
        seqTwoSum = sum(nums[k:k * 2])
        seqThreeSum = sum(nums[k * 2:k * 3])

        # Sums of combined best windows
        bestSeqSum = seqSum
        bestTwoSum = seqSum + seqTwoSum
        bestThreeSum = seqSum + seqTwoSum + seqThreeSum

        # Current window positions
        seqIndex = 1
        twoSeqIndex = k + 1
        threeSeqIndex = k * 2 + 1
        while threeSeqIndex <= len(nums) - k:
            # Update the three sliding windows
            seqSum = seqSum - nums[seqIndex - 1] + nums[seqIndex + k - 1]
            seqTwoSum = seqTwoSum - nums[twoSeqIndex - 1] + nums[twoSeqIndex + k - 1]
            seqThreeSum = seqThreeSum - nums[threeSeqIndex - 1] + nums[threeSeqIndex + k - 1]

            # Update best single window
            if seqSum > bestSeqSum:
                bestSeq = seqIndex
                bestSeqSum = seqSum

            # Update best two windows
            if seqTwoSum + bestSeqSum > bestTwoSum:
                bestTwoSeq = [bestSeq, twoSeqIndex]
                bestTwoSum = seqTwoSum + bestSeqSum

            # Update best three windows
            if seqThreeSum + bestTwoSum > bestThreeSum:
                bestThreeSeq = bestTwoSeq + [threeSeqIndex]
                bestThreeSum = seqThreeSum + bestTwoSum

            # Update the current positions
            seqIndex += 1
            twoSeqIndex += 1
            threeSeqIndex += 1

        return bestThreeSeq


if __name__ == '__main__':
    pass