How do you solve Partition Array for Maximum Sum on LeetCode?

Partition Array for Maximum Sum (LeetCode #1043) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally and efficiently.

What is the brute force solution for Partition Array for Maximum Sum?

The brute force approach for Partition Array for Maximum Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible partition of the array and calculating the maximum sum for each partition. This is done by iterating through the array and considering all subarrays of length up to k. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Partition Array for Maximum Sum?

Partition Array for Maximum Sum uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Partition Array for Maximum Sum?

Always clarify the problem constraints and edge cases before diving into coding. Think aloud during the interview to show your thought process and reasoning. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Partition Array for Maximum Sum?

Failing to consider the bounds of the array when checking subarrays. Not initializing the dp array correctly or misunderstanding its purpose.

#1043

Partition Array for Maximum Sum

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k)
Space	O(1)	O(n)

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Intuition

Time O(n * k)Space O(n)

The optimal solution uses dynamic programming to build the maximum sum iteratively. We maintain a dp array where dp[i] represents the maximum sum we can achieve using the first i elements of the array.

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Algorithm

3 steps

1Step 1: Initialize a dp array of size n+1 with all zeros.
2Step 2: Iterate through the array from 1 to n.
3Step 3: For each position i, check the last j elements (up to k), calculate the maximum value in that segment, and update dp[i] accordingly.

solution.py9 lines

1def maxSumAfterPartitioning(arr, k):
2    n = len(arr)
3    dp = [0] * (n + 1)
4    for i in range(1, n + 1):
5        max_val = 0
6        for j in range(1, min(k, i) + 1):
7            max_val = max(max_val, arr[i - j])
8            dp[i] = max(dp[i], dp[i - j] + max_val * j)
9    return dp[n]

ℹ

Complexity note: The time complexity is O(n * k) because we iterate through each element and for each element, we may check up to k previous elements, leading to a nested loop structure.

1Dynamic programming allows us to build solutions incrementally and efficiently.
2Understanding how to maintain state (like max values) is crucial in optimizing solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.