1043. Partition Array for Maximum Sum

Problem

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

  Constraints:

• 1 <= arr.length <= 500

• 0 <= arr[i] <= 10^9

• 1 <= k <= arr.length

Solution (Java)

class Solution {
    public int maxSumAfterPartitioning(int[] arr, int k) {
        int n = arr.length;
        int[] dp = new int[n];
        for (int right = 0; right < n; right++) {
            int localMax = arr[right];
            for (int left = right; left > Math.max(-1, right - k); left--) {
                localMax = Math.max(localMax, arr[left]);
                if (left == 0) {
                    dp[right] = Math.max(dp[right], (right - left + 1) * localMax);
                } else {
                    dp[right] = Math.max(dp[right], dp[left - 1] + (right - left + 1) * localMax);
                }
            }
        }
        return dp[n - 1];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).