1043. Partition Array for Maximum Sum

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# 題目敘述

Given an integer array arr , partition the array into (contiguous) subarrays of length at most k . After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a 32-bit integer.

# Example 1

Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]

# Example 2

Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83

# Example 3

Input: arr = [1], k = 1
Output: 1

# 解題思路

# Solution

class Solution {

    public int maxSumAfterPartitioning(int[] arr, int k) {

        int dp[] = new int[arr.length + 1];

        for (int index = arr.length - 1; index >= 0; index--) {

            int max = Integer.MIN_VALUE;

            int ans = Integer.MIN_VALUE;

            int len = 0;

            for (int i = index; i < index + k && i < arr.length; i++) {

                len++;

                max = Math.max(max, arr[i]);

                ans = Math.max(ans, len * max + dp[i + 1]);

            }

            dp[index] = ans;

        }

        return dp[0];

    }

}