LeetCode-in-Java

1043. Partition Array for Maximum Sum

Medium

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] <= 109
• 1 <= k <= arr.length

Solution

public 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];
    }
}

This site is open source. Improve this page.