<p>You are given a string <code>s</code> consisting only of characters <code>&#39;a&#39;</code> and <code>&#39;b&#39;</code>​​​​.</p>

<p>You can delete any number of characters in <code>s</code> to make <code>s</code> <strong>balanced</strong>. <code>s</code> is <strong>balanced</strong> if there is no pair of indices <code>(i,j)</code> such that <code>i &lt; j</code> and <code>s[i] = &#39;b&#39;</code> and <code>s[j]= &#39;a&#39;</code>.</p>

<p>Return <em>the <strong>minimum</strong> number of deletions needed to make </em><code>s</code><em> <strong>balanced</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aababbab&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can either:
Delete the characters at 0-indexed positions 2 and 6 (&quot;aa<u>b</u>abb<u>a</u>b&quot; -&gt; &quot;aaabbb&quot;), or
Delete the characters at 0-indexed positions 3 and 6 (&quot;aab<u>a</u>bb<u>a</u>b&quot; -&gt; &quot;aabbbb&quot;).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bbaaaaabb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only solution is to delete the first two characters.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is&nbsp;<code>&#39;a&#39;</code> or <code>&#39;b&#39;</code>​​.</li>
</ul>

/*Approach 4: Using stack (one pass)
Intuition
What if we focus on removing "ba" pairs? These pairs unbalance the string because an 'a' character is to the right of a 'b' character. 
By leveraging a stack, we can efficiently count and remove these pairs in a single traversal of the string.

To implement this approach, we traverse the string and push each character onto the stack. When we encounter a "ba" pair—where an 'a' is 
on top of the stack and a 'b' is currently being processed—we pop the 'a' from the stack, effectively "removing" this out-of-order pair. 
We keep a count of such removals throughout this process.

However, in the worst case (when no deletions are needed), it still uses O(n) space for the stack.
*/
class Solution {
public:
    int minimumDeletions(string s) {
        int n = s.length();
        stack<char> charStack;
        int deleteCount = 0;

        // Iterate through each character in the string
        for (int i = 0; i < n; i++) {
            // If stack is not empty, top of stack is 'b',
            // and current char is 'a'
            if (!charStack.empty() && charStack.top() == 'b' && s[i] == 'a') {
                charStack.pop();  // Remove 'b' from stack
                deleteCount++;    // Increment deletion count
            } else {
                charStack.push(s[i]);  // Push current character onto stack
            }
        }

        return deleteCount;
    }
};
/*Complexity Analysis
Let n be the size of string s.

Time complexity: O(n)
The algorithm performs a single linear pass over the string, with stack operations (push and pop) taking O(1) time.

Space complexity: O(n)
The algorithm uses a stack that may grow up to the size of the string.
*/
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