How do you solve Remove K-Balanced Substrings on LeetCode?

Remove K-Balanced Substrings (LeetCode #3703) 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: Using a stack helps manage nested structures efficiently.

What is the time complexity of Remove K-Balanced Substrings?

The optimal solution for Remove K-Balanced Substrings runs in O(n) time and uses O(n) space. The linear time complexity comes from a single traversal of the string, while space is used to store the balance in the stack.

What is the brute force solution for Remove K-Balanced Substrings?

The brute force approach for Remove K-Balanced Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach repeatedly scans the string to find and remove all k-balanced substrings until no more can be found. It’s straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove K-Balanced Substrings?

Remove K-Balanced Substrings uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Remove K-Balanced Substrings?

Explain your thought process clearly. Consider edge cases like empty strings or no balanced substrings. Practice similar problems to recognize patterns.

What are common mistakes when solving Remove K-Balanced Substrings?

Not accounting for nested k-balanced substrings. Incorrectly managing the stack size during removals.

#3703

Remove K-Balanced Substrings

Medium

StringStackSimulationStackGreedy

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)

Using a stack to track the balance of parentheses allows us to efficiently identify and remove k-balanced substrings in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Use a stack to track counts of '(' and ')' and identify k-balanced sections.
2Step 2: For every k pairs found, mark them for removal.
3Step 3: Construct the final string excluding the marked sections.

solution.py7 lines

1def removeKbalanced(s, k):
2    stack = []
3    for char in s:
4        if char == '(': stack.append(1)
5        elif char == ')' and len(stack) >= k:
6            for _ in range(k): stack.pop()
7    return ''.join(['(' * len(stack) + ')' * len(stack)])

ℹ

Complexity note: The linear time complexity comes from a single traversal of the string, while space is used to store the balance in the stack.

1Using a stack helps manage nested structures efficiently.
2Identifying patterns of parentheses allows for targeted removals.

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