[LeetCode] 689. Maximum Sum of 3 Non-Overlapping Subarrays

[grandyang]

 

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).

 

这道题给了我们一个只包含正数的数组，让我们找三个长度为k的不重叠的子数组，使得所有子数组的数字之和最大。首先我们应该明确的是，暴力搜索在这道题上基本不太可能，因为遍历一个子数组的复杂度是平方级，遍历三个还不得六次方啊，看OJ不削你～那么我们只能另辟蹊径，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，其中:

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，参见代码如下：

 

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);
        for (int num : nums) sums.push_back(sums.back() + num);
        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];
            }
        }
        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];
            }
        }
        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;
    }
};

 

类似题目：

Best Time to Buy and Sell Stock III

 

参考资料：

https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/discuss/108231/C++Java-DP-with-explanation-O(n)

https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/discuss/108246/C++-O(n)-time-O(n)-space-concise-solution

https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/discuss/108230/Clean-Java-DP-O(n)-Solution.-Easy-extend-to-Sum-of-K-Non-Overlapping-SubArrays

 

LeetCode All in One 题目讲解汇总(持续更新中...)