How do you solve Maximize Active Section with Trade I on LeetCode?

Maximize Active Section with Trade I (LeetCode #3499) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying segments helps optimize the problem.

What is the time complexity of Maximize Active Section with Trade I?

The optimal solution for Maximize Active Section with Trade I runs in O(n) time and uses O(n) space. Linear time complexity due to single pass through the string and segments.

What is the brute force solution for Maximize Active Section with Trade I?

The brute force approach for Maximize Active Section with Trade I is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible trades by iterating through the string and calculating the active sections after each trade. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize Active Section with Trade I?

Maximize Active Section with Trade I uses the following concepts: String, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximize Active Section with Trade I?

Clarify the problem statement and constraints. Discuss trade-offs between brute force and optimal solutions. Practice similar string manipulation problems.

What are common mistakes when solving Maximize Active Section with Trade I?

Not considering edge cases with no '1's. Overlooking the need for surrounding '0's.

#3499

Maximize Active Section with Trade I

Medium

StringEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Count segments of '1's and '0's, then calculate potential trades efficiently by considering only valid segments.

⚙️

Algorithm

3 steps

1Step 1: Split the string into segments of '1's and '0's.
2Step 2: For each block of '1's surrounded by '0's, calculate the potential increase in active sections.
3Step 3: Return the maximum active sections after considering all valid trades.

solution.py9 lines

1def maxActiveSections(s):
2    s = '1' + s + '1'
3    segments = [len(x) for x in s.split('0')]
4    max_active = segments.count(1)
5    for i in range(1, len(segments) - 1):
6        if segments[i] > 1:
7            new_active = max_active - 1 + 2
8            max_active = max(max_active, new_active)
9    return max_active

ℹ

Complexity note: Linear time complexity due to single pass through the string and segments.

1Identifying segments helps optimize the problem.
2Surrounding conditions dictate valid trades.

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