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 = 3Output: 84Explanation: arr becomes [15,15,15,9,10,10,10]
Question
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 = 3Output: 84Explanation: arr becomes [15,15,15,9,10,10,10]
Solution
Sure, here is a Python solution for the problem:
def maxSumAfterPartitioning(arr, k):
dp = [0] * (len(arr) + 1)
for i in range(1, len(arr) + 1):
max_val = 0
for j in range(1, min(i, k) + 1):
max_val = max(max_val, arr[i - j])
dp[i] = max(dp[i], dp[i - j] + max_val * j)
return dp[-1]
This function uses dynamic programming to solve the problem. The dp array stores the maximum sum we can get for the first i elements of the array. For each i, we try all possible lengths of the last subarray (from 1 to min(i, k)), update the maximum value in the last subarray (max_val), and update dp[i] with the maximum sum we can get.
For example, if arr = [1,15,7,9,2,5,10] and k = 3, the function will return 84.
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