How do you solve Minimum Deletions to Make String Balanced on LeetCode?

Minimum Deletions to Make String Balanced (LeetCode #1653) 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: The order of characters matters; 'b's must always come before 'a's to maintain balance.

What is the brute force solution for Minimum Deletions to Make String Balanced?

The brute force approach for Minimum Deletions to Make String Balanced is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can check all possible substrings of the input string and count how many deletions are needed to make each substring balanced. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Deletions to Make String Balanced?

Minimum Deletions to Make String Balanced uses the following concepts: String, Dynamic Programming, Stack. The recommended patterns to study are: Two Pointers, Greedy Algorithms.

What are the interview tips for Minimum Deletions to Make String Balanced?

Always start with a brute-force solution to understand the problem before optimizing. Think about how you can reduce the number of operations by keeping track of counts instead of manipulating strings. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Minimum Deletions to Make String Balanced?

Not considering the order of characters when counting deletions. Overcomplicating the problem by trying to generate all substrings instead of focusing on a linear scan.

#1653

Minimum Deletions to Make String Balanced

Medium

StringDynamic ProgrammingStackTwo PointersGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages a single pass through the string to count the number of 'b's before each 'a' and calculates the minimum deletions needed in a more efficient manner. This approach is faster and scales better with larger strings.

⚙️

Algorithm

3 steps

1Step 1: Initialize two counters: one for the count of 'b's seen so far and another for the total deletions needed.
2Step 2: Traverse the string from left to right. For each character: - If it's 'b', increment the count of 'b's. - If it's 'a', add the count of 'b's to the deletions needed.
3Step 3: The final count of deletions will give the minimum deletions needed to make the string balanced.

solution.py14 lines

1# Full working Python code
2
3def min_deletions_optimal(s):
4    b_count = 0
5    deletions = 0
6    for char in s:
7        if char == 'b':
8            b_count += 1
9        elif char == 'a':
10            deletions += b_count
11    return deletions
12
13# Example usage:
14print(min_deletions_optimal('aababbab'))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string. The space complexity is O(1) since we only use a constant amount of space for counters.

1The order of characters matters; 'b's must always come before 'a's to maintain balance.
2Counting occurrences of 'b's before each 'a' helps determine the necessary deletions efficiently.

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