# 1043. Partition Array for Maximum Sum

## Description

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

## Solutions

Solution 1: Dynamic Programming

We define $f[i]$ to represent the maximum element sum of the first $i$ elements of the array after separating them into several subarrays. At the beginning, $f[i]=0$, and the answer is $f[n]$.

We consider how to calculate $f[i]$, where $i \geq 1$.

For $f[i]$, its last element is $arr[i-1]$. Since the maximum length of each subarray is $k$, and we need to find the maximum value in the subarray, we can enumerate the first element $arr[j - 1]$ of the last subarray from right to left, where $\max(0, i - k) \lt j \leq i$, and maintain a variable $mx$ during the process to represent the maximum value in the subarray. The state transition equation is:

$f[i] = \max\{f[i], f[j - 1] + mx \times (i - j + 1)\}$

The final answer is $f[n]$.

The time complexity is $O(n \times k)$, and the space complexity is $O(n)$, where $n$ is the length of the array $arr$.

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

• class Solution {
public:
int maxSumAfterPartitioning(vector<int>& arr, int k) {
int n = arr.size();
int f[n + 1];
memset(f, 0, sizeof(f));
for (int i = 1; i <= n; ++i) {
int mx = 0;
for (int j = i; j > max(0, i - k); --j) {
mx = max(mx, arr[j - 1]);
f[i] = max(f[i], f[j - 1] + mx * (i - j + 1));
}
}
return f[n];
}
};

• class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
f = [0] * (n + 1)
for i in range(1, n + 1):
mx = 0
for j in range(i, max(0, i - k), -1):
mx = max(mx, arr[j - 1])
f[i] = max(f[i], f[j - 1] + mx * (i - j + 1))
return f[n]


• func maxSumAfterPartitioning(arr []int, k int) int {
n := len(arr)
f := make([]int, n+1)
for i := 1; i <= n; i++ {
mx := 0
for j := i; j > max(0, i-k); j-- {
mx = max(mx, arr[j-1])
f[i] = max(f[i], f[j-1]+mx*(i-j+1))
}
}
return f[n]
}

• function maxSumAfterPartitioning(arr: number[], k: number): number {
const n: number = arr.length;
const f: number[] = new Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
let mx: number = 0;
for (let j = i; j > Math.max(0, i - k); --j) {
mx = Math.max(mx, arr[j - 1]);
f[i] = Math.max(f[i], f[j - 1] + mx * (i - j + 1));
}
}
return f[n];
}