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# Partition Array For Maximum Sum - Leetcode Solution

LeetCode: Partition Array For Maximum Sum Leetcode Solution

Difficulty: Medium

Topics: dynamic-programming array

The Partition Array for Maximum Sum problem on LeetCode is a dynamic programming problem that requires us to partition an array into subarrays, with the goal of maximizing the sum of each subarray. Here's a detailed solution:

First, we create a new array dp of length n, where n is the length of the input array A. Each element dp[i] will represent the maximum sum we can obtain from partitioning the subarray A[0:i]. We initialize dp[0] to be A[0] (since the subarray containing only the first element is just the first element itself).

Then, for each index i (1 <= i < n), we consider all possible sizes j of the last subarray, where j ranges from 1 to K (the maximum partition size). For each such j, we calculate the sum of the last subarray A[i-j+1:i+1], and add this sum to the maximum sum we can obtain from partitioning the rest of the array A[0:i-j]. This gives us a candidate value of dp[i] for this particular j value.

We keep track of the maximum value of dp[i] over all possible j values, and set dp[i] to be this maximum value. We repeat this process for all i, until we reach the end of the input array A.

Finally, the maximum sum we can obtain from partitioning the entire array A is simply the value of dp[n-1], since dp[n-1] represents the maximum sum we can obtain from partitioning the subarray A[0:n-1].

Here's the full Python code implementation:

def maxSumAfterPartitioning(A, K): n = len(A) dp = [0] * n dp[0] = A[0]

```
for i in range(1, n):
max_val = A[i]
for j in range(1, min(i+1, K+1)):
max_val = max(max_val, A[i-j+1])
if i >= j:
dp[i] = max(dp[i], dp[i-j] + max_val*j)
else:
dp[i] = max(dp[i], max_val*j)
return dp[n-1]
```

We iterate over all indices i in the input array A, and for each index, we consider all possible last subarray sizes j (up to K). We then update the maximum sum we can obtain for the current index i by considering all previous indices (up to i-j) and adding the sum of the last subarray to it. We take the maximum value of this across all possible last subarray sizes j, and set dp[i] equal to it. Finally, we return dp[n-1], which represents the maximum sum we can obtain from partitioning the entire array A.

## Partition Array For Maximum Sum Solution Code

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