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

Maximize Active Section with Trade II (LeetCode #3501) 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: Identify segments of '1's and '0's.

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

The optimal solution for Maximize Active Section with Trade II runs in O(n) time and uses O(n) space. The complexity is linear since we only traverse the string and segments once, making it efficient.

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

The brute force approach for Maximize Active Section with Trade II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible trade for each query, which is straightforward but inefficient. This method involves simulating the trade for all segments in the substring. The optimal Optimal Solution approach improves this to O(n).

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

Maximize Active Section with Trade II uses the following concepts: Array, String, Binary Search, Segment Tree. The recommended patterns to study are: Hash Map, Array.

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

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice segment identification problems.

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

Not handling edge cases with boundaries. Overlooking the need to augment the string with '1's.

#3501

Maximize Active Section with Trade II

Hard

ArrayStringBinary SearchSegment TreeHash 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)

Identify segments of '1's and '0's, then calculate the maximum active sections by merging segments intelligently. This avoids redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Parse the string into segments of '1's and '0's, assigning IDs to each segment.
2Step 2: For each query, check the segments around the trade position and calculate the potential new active sections.
3Step 3: Return the maximum possible active sections after the optimal trade.

solution.py22 lines

1def maxActiveSections(s, queries):
2    segments = []
3    count = 0
4    for char in s:
5        if char == '1':
6            if count >= 0:
7                count += 1
8            else:
9                segments.append(count)
10                count = 1
11        else:
12            if count <= 0:
13                count -= 1
14            else:
15                segments.append(count)
16                count = -1
17    segments.append(count)
18    results = []
19    for l, r in queries:
20        max_count = segments[l - 1] + segments[r + 1] if l > 0 and r < len(segments) - 1 else 0
21        results.append(max_count)
22    return results

ℹ

Complexity note: The complexity is linear since we only traverse the string and segments once, making it efficient.

1Identify segments of '1's and '0's.
2Merging segments intelligently maximizes active sections.

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