1043. Partition Array for Maximum Sum

Link

Given an integer array arr, you should 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.

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

Constraints:

• 1 <= arr.length <= 500

• 0 <= arr[i] <= 109

• 1 <= k <= arr.length

Solution

dp[i] : the max result of arr[0....i]

consider replacing the last k element of arr[0...i] to the max value among them, to see if they can derive the max result:

dp[i] = Math.max(dp[i], dp[i-k] + k*curMax);

class Solution {
    public int maxSumAfterPartitioning(int[] arr, int k) {
        int n = arr.length;
        int[] dp = new int[n];
        int curMax;
        for(int i = 0; i < n; i++){
            curMax = 0;
            for(int j = 1; (j <= k)&&(i-j+1 >= 0); j++){
                curMax = Math.max(curMax, arr[i-j+1]);
                dp[i] = Math.max(dp[i], (i-j >= 0 ? dp[i-j] : 0) + curMax*j);
            }
        }
        return dp[n-1];
        
    }
}