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Formatted question description: https://leetcode.ca/all/1043.html

# 1043. Partition Array for Maximum Sum (Medium)

Given an integer array A, you 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: A = [1,15,7,9,2,5,10], K = 3
Output: 84
Explanation: A becomes [15,15,15,9,10,10,10]

Note:

1. 1 <= K <= A.length <= 500
2. 0 <= A[i] <= 10^6

Related Topics:
Graph

## Solution 1. DP

Let dp[i] be the subproblem in A[0..i].

dp[i] = max( dp[j-1] + max(A[j], ..., A[i]) * (i - j + 1) | i - j + 1 <= K && j >= 0 )
dp[-1] = 0

// OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Time: O(NK)
// Space: O(N)
class Solution {
public:
int maxSumAfterPartitioning(vector<int>& A, int K) {
int N = A.size();
vector<int> dp(N);
for (int i = 0; i < N; ++i) {
int maxVal = 0;
for (int j = i, k = 0; j >= 0 && k < K; --j, ++k) {
maxVal = max(maxVal, A[j]);
dp[i] = max(dp[i], (j == 0 ? 0 : dp[j - 1]) + maxVal * (i - j + 1));
}
}
return dp[N - 1];
}
};


## Solution 2. DP

// OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Time: O(NK)
// Space: O(K)
class Solution {
public:
int maxSumAfterPartitioning(vector<int>& A, int K) {
int N = A.size();
++K;
vector<int> dp(K);
for (int i = 0; i < N; ++i) {
int maxVal = 0;
dp[i % K] = 0;
for (int j = i, k = 0; j >= 0 && k < K - 1; --j, ++k) {
maxVal = max(maxVal, A[j]);
dp[i % K] = max(dp[i % K], (j == 0 ? 0 : dp[(j - 1) % K]) + maxVal * (i - j + 1));
}
}
return dp[(N - 1) % K];
}
};

• class Solution {
public int maxSumAfterPartitioning(int[] A, int K) {
if (A == null || A.length == 0)
return 0;
int length = A.length;
int[] dp = new int[length + 1];
for (int i = 1; i <= length; i++) {
int max = 0;
int leftBound = Math.max(i - K, 0);
for (int j = i - 1; j >= leftBound; j--) {
max = Math.max(max, A[j]);
dp[i] = Math.max(dp[i], dp[j] + max * (i - j));
}
}
return dp[length];
}
}

############

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

• // OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Time: O(NK)
// Space: O(N)
class Solution {
public:
int maxSumAfterPartitioning(vector<int>& A, int K) {
int dp[501] = {}, N = A.size();
for (int i = 0; i < N; ++i) {
int mx = 0;
for (int t = i, last = max(0, i + 1 - K); t >= last; --t) {
mx = max(mx, A[t]);
dp[i + 1] = max(dp[i + 1], dp[t] + mx * (i - t + 1));
}
}
return dp[N];
}
};

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

############

# 1043. Partition Array for Maximum Sum
# https://leetcode.com/problems/partition-array-for-maximum-sum/

class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
dp = [0] * (N + 1)

for i in range(1, N + 1):
mmax = 0
for k in range(1, min(i, K) + 1):
mmax = max(mmax, A[i - k])
dp[i] = max(dp[i], dp[i - k] + mmax * k)

return dp[N]


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

func max(a, b int) int {
if a > b {
return a
}
return b
}

• 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];
}