689. Maximum Sum of 3 Non-Overlapping Subarrays

Description

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

Each subarray will be of sizek, and we want to maximize the sum of all3*kentries.

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.lengthwill be between 1 and 20000.

nums[i]will be between 1 and 65535.

kwill be between 1 and floor(nums.length / 3).

Discussion

Method 1 O(n^2)

区间dp[i][j] + dfs, 从选择从i-j 的最大值。

Method 2 O(n) Leet Code

构建left[], right[].

left[i], from 0 - i, the start index of max chunk

right[i], from i to n, the start index of max chunk.

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size(), maxsum = 0;
        vector<int> sum = {0}, posLeft(n, 0), posRight(n, n-k), ans(3, 0);
        for (int i:nums) sum.push_back(sum.back()+i);
       // DP for starting index of the left max sum interval
        for (int i = k, tot = sum[k]-sum[0]; i < n; i++) {
            if (sum[i+1]-sum[i+1-k] > tot) {
                posLeft[i] = i+1-k;
                tot = sum[i+1]-sum[i+1-k];
            }
            else 
                posLeft[i] = posLeft[i-1];
        }
        // DP for starting index of the right max sum interval
        // caution: the condition is ">= tot" for right interval, and "> tot" for left interval
        for (int i = n-k-1, tot = sum[n]-sum[n-k]; i >= 0; i--) {
            if (sum[i+k]-sum[i] >= tot) {
                posRight[i] = i;
                tot = sum[i+k]-sum[i];
            }
            else
                posRight[i] = posRight[i+1];
        }
        // test all possible middle interval
        for (int i = k; i <= n-2*k; i++) {
            int l = posLeft[i-1], r = posRight[i+k];
            int tot = (sum[i+k]-sum[i]) + (sum[l+k]-sum[l]) + (sum[r+k]-sum[r]);
            if (tot > maxsum) {
                maxsum = tot;
                ans = {l, i, r};
            }
        }
        return ans;
    }
};