How do you solve Minimum String Length After Balanced Removals on LeetCode?

Minimum String Length After Balanced Removals (LeetCode #3746) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Removing balanced substrings reduces the string length.

What is the time complexity of Minimum String Length After Balanced Removals?

The optimal solution for Minimum String Length After Balanced Removals runs in O(n) time and uses O(1) space. Single pass through the string gives O(n) complexity with O(1) space.

What is the brute force solution for Minimum String Length After Balanced Removals?

The brute force approach for Minimum String Length After Balanced Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible substrings and remove those that are balanced. This approach is straightforward but inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum String Length After Balanced Removals?

Minimum String Length After Balanced Removals uses the following concepts: String, Stack, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum String Length After Balanced Removals?

Explain your thought process clearly. Consider edge cases like empty strings. Discuss time and space complexity.

What are common mistakes when solving Minimum String Length After Balanced Removals?

Not counting characters correctly. Assuming all characters can be removed.

#3746

Minimum String Length After Balanced Removals

Medium

StringStackCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Count the number of 'a's and 'b's. The minimum length is determined by the absolute difference between these counts, as unbalanced characters cannot be removed.

⚙️

Algorithm

3 steps

1Step 1: Count the occurrences of 'a' and 'b' in the string.
2Step 2: Calculate the absolute difference between the counts.
3Step 3: Return the difference as the minimum length.

solution.py4 lines

1def min_length_optimal(s):
2    count_a = s.count('a')
3    count_b = s.count('b')
4    return abs(count_a - count_b)

ℹ

Complexity note: Single pass through the string gives O(n) complexity with O(1) space.

1Removing balanced substrings reduces the string length.
2The final length depends only on the unbalanced characters.

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