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] <= 109

• 1 <= k <= arr.length

Solution

/**
 * @param {number[]} arr
 * @param {number} k
 * @return {number}
 */
var maxSumAfterPartitioning = function(arr, k) {
    var dp = Array(arr.length).fill(0);
    for (var i = 0; i < arr.length; i++) {
        var maxValue = 0;
        for (var j = 1; j <= k && j - 1 <= i; j++) {
            maxValue = Math.max(maxValue, arr[i - j + 1]);
            dp[i] = Math.max(dp[i], ( dp[i - j + 1] + maxValue * j);
        }
    }
    return dp[arr.length];
};

Explain:

dp[i + 1] represents array [0 ... i] 's maxSumAfterPartitioning. (dp[0] = 0 means nothing, it just helps calculate.)

dp[i] = max(
  dp[i - 1] + max(arr[i]) * 1,
  dp[i - 2] + max(arr[i], arr[i - 1]) * 2,
  ...
  dp[i - k] + max(arr[i], arr[i - 1], ..., arr[i - k]) * k
)

dp[arr.length - 1] would be the answer in the end of the iteration.

Complexity:

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