Question

题目大意：给定正整数数组nums，计算其中不想交的 3 段子数组的最大和。每段子数组的长度为k。

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.

Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.

Example:

Input: [1,2,1,2,6,7,5,1], 2 Output: [0, 3, 5] Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5]. We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.

Note:

• nums.length will be between 1 and 20000.
• nums[i] will be between 1 and 65535.
• k will be between 1 and floor(nums.length / 3).

Difficulty:Hard

Category: Array, DP

Analyze

解法来源于博客：Maximum Sum of 3 Non-Overlapping Subarrays 三个非重叠子数组的最大和

对于这种求子数组和有关的题目时，一般都需要建立累加和数组，为啥呢，因为累加和数组可以快速的求出任意长度的子数组之和，当然也能快速的求出长度为k的子数组之和。

因为这道题只让我们找出三个子数组，那么我们可以先确定中间那个子数组的位置，这样左右两边的子数组的位置范围就缩小了，中间子数组的起点不能是从开头到结尾整个区间，必须要在首尾各留出k个位置给其他两个数组。一旦中间子数组的起始位置确定了，那么其和就能通过累加和数组快速确定。那么现在就要在左右两边的区间内分别找出和最大的子数组，遍历所有的子数组显然不是很高效，如何快速求出呢，这里我们需要使用动态规划Dynamic Programming的思想来维护两个DP数组left和right.

• left[i]表示在区间[0, i]范围内长度为k且和最大的子数组的起始位置
• right[i]表示在区间[i, n - 1]范围内长度为k且和最大的子数组的起始位置

这两个dp数组各需要一个for循环来更新，left数组都初始化为0，前k个数字没办法，肯定起点都是0，变量total初始化为前k个数字之和，然后从第k+1个数字开始，每次向前取k个，利用累加和数组sums快速算出数字之和，跟total比较，如果大于total的话，那么更新total和left数组当前位置值，否则的话left数组的当前值就赋值为前一位的值。同理对right数组的更新也类似，total初始化为最后k个数字之和，然后从前一个数字向前遍历，如果大于total，更新total和right数组的当前位置，否则right数组的当前值就赋值为后一位的值。一旦left数组和right数组都更新好了，那么就可以遍历中间子数组的起始位置了，然后我们可以通过left和right数组快速定位出左边和右边的最大子数组的起始位置，并快速计算出这三个子数组的所有数字之和，用来更新全局最大值mx，如果mx被更新了的话，记录此时的三个子数组的起始位置到结果res中

Solution

Time complexity: O(n) Space complexity: O(n)

class Solution {
 public:
  vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
    int n = nums.size(), mx = INT_MIN;
    vector<int> sums{0}, res, left(n, 0), right(n, n - k);

    // 2-DP: left, right
    // left[i] == the start index with len k which have the biggest sum
    // right[i] == the start index with len k which have the biggeest sum
    // Get the cumulative sum
    for (int num : nums) sums.push_back(sums.back() + num);

    // Pretreatment for the left dp
    for (int i = k, total = sums[k] - sums[0]; i < n; ++i) {
      if (sums[i + 1] - sums[i + 1 - k] > total) {
        left[i] = i + 1 - k;
        total = sums[i + 1] - sums[i + 1 - k];
      } else {
        left[i] = left[i - 1];
      }
    }

    // Pretreatment for the right dp
    for (int i = n - 1 - k, total = sums[n] - sums[n - k]; i >= 0; --i) {
      if (sums[i + k] - sums[i] >= total) {
        right[i] = i;
        total = sums[i + k] - sums[i];
      } else {
        right[i] = right[i + 1];
      }
    }

    // Main function
    for (int i = k; i <= n - 2 * k; ++i) {
      int l = left[i - 1], r = right[i + k];
      int total = (sums[i + k] - sums[i]) + (sums[l + k] - sums[l]) + (sums[r + k] - sums[r]);
      if (mx < total) {
        mx = total;
        res = {l, i, r};
      }
    }
    return res;
  }
};