# Solving Leetcode Interviews in Seconds with AI: Partition Array for Maximum Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1043" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming:** We use dynamic programming to store and reuse the maximum sum achievable up to each index. `dp[i]` stores the largest sum after partitioning `arr[0...i]`.

*   **Iterative Calculation:** For each index `i`, we iterate back `j` positions (up to `k`) to consider all possible subarray lengths ending at `i`. We find the maximum value within that subarray and calculate the potential sum by adding it to the previously computed optimal sum `dp[i-j]`.

*   **Maximization:** We update `dp[i]` with the maximum sum found among all possible subarray lengths ending at `i`.

*   **Time & Space Complexity**: O(n*k) time complexity, O(n) space complexity.

	
	# Code
	```python
	def maxSumAfterPartitioning(arr: list[int], k: int) -> int:
    """
    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.

    For example:
    maxSumAfterPartitioning([1,15,7,9,2,5,10], 3) == 84
    maxSumAfterPartitioning([1,4,1,5,7,3,6,1,9,9,3], 4) == 83
    maxSumAfterPartitioning([1], 1) == 1
    """
    n = len(arr)
    dp = [0] * n
    for i in range(n):
        max_val = 0
        for j in range(1, min(k, i + 1) + 1):
            max_val = max(max_val, arr[i - j + 1])
            if i - j >= -1:
                if i - j == -1:
                    dp[i] = max(dp[i], max_val * j)
                else:
                    dp[i] = max(dp[i], dp[i - j] + max_val * j)
    return dp[n - 1]
	```