LeetCode 1043. Separate the array to get the maximum and daily questions

Problem description

Give you an array of integers arr, Please separate the array into at most k Some of （ continuity ） Subarray . After separation , All values in each subarray will become the maximum value in that subarray .

Returns the maximum sum of elements that can be obtained after separating and transforming the array .

Be careful , The corresponding order of the original array and the separated array should be the same , in other words , You can only choose the position of the separated array, but you can't adjust the order in the array .

Example 1：

Input ：arr = [1,15,7,9,2,5,10], k = 3
Output ：84
explain ：
because k=3 Can be separated into [1,15,7] [9] [2,5,10], The result is [15,15,15,9,10,10,10], And for 84, Is the largest sum of all elements of the array after separation and transformation .
If separated into [1] [15,7,9] [2,5,10], The result is [1, 15, 15, 15, 10, 10, 10] But the sum of the elements of this separation （76） Less than one . 
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
 

Tips ：

1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/partition-array-for-maximum-sum
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Java

class Solution {
    public int maxSumAfterPartitioning(int[] arr, int k) {
        int n = arr.length;
        int[] dp = new int[n];
        for(int i = 0;i < k;i++){
            dp[i] = getMaxValue(arr,0,i) * (i + 1);
        }
        for(int i = k;i < n;i++){
            for(int j = 1;j <= k;j++){
                dp[i] = Math.max(dp[i],dp[i - j] + getMaxValue(arr,i - j + 1,i) * j);
            }
        }
        return dp[n - 1];
    }
    // Get the maximum value of the specified interval 
    public int getMaxValue(int[] arr,int left,int right){
        int ans = 0;
        for(int i = left;i <= right;i++){
            ans = Math.max(ans,arr[i]);
        }
        return ans;
    }
}