# Solving Leetcode Interviews in Seconds with AI: Minimum Deletions to Make String Balanced


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1653" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a string s consisting only of characters 'a' and 'b'​​​​. You can delete any number of characters in s to make s balanced. s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = 'b' and s[j]= 'a'. Return the minimum number of deletions needed to make s balanced.   Example 1:  Input: s = "aababbab" Output: 2 Explanation: You can either: Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").  Example 2:  Input: s = "bbaaaaabb" Output: 2 Explanation: The only solution is to delete the first two characters.    Constraints:  1 <= s.length <= 105 s[i] is 'a' or 'b'​​.  

	# Explanation
	Here's the breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Iterate through the string, maintaining a count of 'a's encountered so far.
    *   For each 'b' encountered, we have a choice: either delete the 'b' or delete all the 'a's seen before it. We choose the option that results in fewer deletions.
    *   The minimum number of deletions needed will be the result.

*   **Complexity:**
    *   Runtime Complexity: O(n)
    *   Storage Complexity: O(1)

	
	# Code
	```python
	def minDeletions(s: str) -> int:
    """
    Calculates the minimum number of deletions needed to make a string balanced.

    Args:
        s: The input string consisting of 'a' and 'b' characters.

    Returns:
        The minimum number of deletions needed to make the string balanced.
    """

    a_count = 0
    deletions = 0

    for char in s:
        if char == 'a':
            a_count += 1
        else:  # char == 'b'
            deletions = min(deletions + 1, a_count)

    return deletions
	```