How do you solve Maximum Strength of K Disjoint Subarrays on LeetCode?

Maximum Strength of K Disjoint Subarrays (LeetCode #3077) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: The strength formula alternates signs based on the position of the subarray, which can lead to different optimal selections.

What is the brute force solution for Maximum Strength of K Disjoint Subarrays?

The brute force approach for Maximum Strength of K Disjoint Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of k disjoint subarrays and calculating their strengths. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Maximum Strength of K Disjoint Subarrays?

Maximum Strength of K Disjoint Subarrays uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Maximum Strength of K Disjoint Subarrays?

Clearly explain your thought process and how you plan to approach the problem. Be careful with edge cases, especially with negative numbers and small arrays. Practice dynamic programming problems to become familiar with the state transition and table filling.

What are common mistakes when solving Maximum Strength of K Disjoint Subarrays?

Failing to consider the disjoint condition when selecting subarrays. Not correctly implementing the strength calculation based on the subarray indices.

#3077

Maximum Strength of K Disjoint Subarrays

Hard

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal approach uses dynamic programming to efficiently compute the maximum strength by breaking the problem into smaller subproblems and storing intermediate results. This avoids redundant calculations and speeds up the process significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j][x] represents the maximum strength using the first i elements to select j subarrays, with x indicating if the last element is included in the last subarray.
2Step 2: Iterate through the array and update the DP table based on the sums of subarrays and the strength formula.
3Step 3: Return the maximum value from the last row of the DP table for k subarrays.

solution.py16 lines

1# Full working Python code
2import numpy as np
3
4def maxStrength(nums, k):
5    n = len(nums)
6    dp = np.full((n + 1, k + 1, 2), float('-inf'))
7    dp[0][0][0] = 0
8    dp[0][0][1] = 0
9
10    for i in range(1, n + 1):
11        for j in range(1, min(k, i) + 1):
12            for x in range(2):
13                dp[i][j][0] = max(dp[i][j][0], dp[i - 1][j][0])
14                dp[i][j][1] = max(dp[i][j][1], dp[i - 1][j - 1][0] + nums[i - 1])
15                dp[i][j][1] = max(dp[i][j][1], dp[i - 1][j][1] + nums[i - 1])
16    return max(dp[n][k])

ℹ

Complexity note: The time complexity is O(n * k) because we iterate through the array and for each element, we update the DP table for each possible number of subarrays up to k.

1The strength formula alternates signs based on the position of the subarray, which can lead to different optimal selections.
2Dynamic programming allows us to build on previously computed results, making the solution more efficient.

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