Leetcode 689. Maximum Sum of 3 Non-Overlapping Subarrays

Problem:

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).

Solution:

　　这道题算是Hard题中的简单题了，思路也很明确，left[i]表示以小于等于i为开始位置的长度为k的最大子数组的起始位置，right[i]表示以大于等于i为开始位置的长度为k的最大子数组的起始位置。然后我们从中间的子数组开始遍历，根据left数组找到左侧的最大子数组起始位置，根据right数组找到右侧最大子数组起始位置，计算三个子数组之和即可。边界情况处理好就行了。

Code:

 

class Solution {
public:
    int getSum(vector<int> &sum,int start,int end){
        return sum[end+1]-sum[start];
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int m = nums.size();
        vector<int> sum(m+1,0);
        for(int i = 0;i != m;++i){
            sum[i+1] = sum[i]+nums[i];
        }
        vector<int> left(m-k+1);
        int maximal = 0;
        for(int i = k-1;i != m;++i){
            int s = getSum(sum,i-k+1,i);
            if(s > maximal){
                maximal = s;
                left[i-k+1] = i-k+1;
            }
            else
                left[i-k+1] = left[i-k];
        }
        vector<int> right(m-k+1);
        maximal = 0;
        for(int i = m-k;i >= 0;--i){
            int s = getSum(sum,i,i+k-1);
            if(s > maximal){
                maximal = s;
                right[i] = i;
            }
            else
                right[i] = right[i+1];
        }
        maximal = 0;
        vector<int> result(3);
        for(int i = k;i <= m-2*k;++i){
            int l = left[i-k];
            int r = right[i+k];
            int s = getSum(sum,l,l+k-1)+getSum(sum,i,i+k-1)+getSum(sum,r,r+k-1);
            if(s > maximal){
                maximal = s;
                result[0] = l;
                result[1] = i;
                result[2] = r;
            }
        }
        return result;
    }
};