<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>

class Solution {
public:
    int func(int left, int right, string &s, int &a, int &b) {
        if (left >= right) return 0;

        // Remove leading 'a's and trailing 'b's
        while (left <= right && s[left] == 'a') {
            a--;
            left++;
        }
        while (left <= right && s[right] == 'b') {
            b--;
            right--;
        }
        int currentCost = min(a, b);
        // Calculate minimum deletions needed
        if (left <= right) {
            // Cost of removing all 'b's up to the current index
            // and all 'a's after the current index
            
            int leftCountB = 0, rightCountA = 0;

            // Count consecutive 'b's from the left
            while (left <= right && s[left] == 'b') {
                leftCountB++;
                left++;
            }

            // Count consecutive 'a's from the right
            while (left <= right && s[right] == 'a') {
                rightCountA++;
                right--;
            }

            a -= rightCountA;
            b -= leftCountB;

            // Update the cost
            currentCost = min(currentCost, leftCountB + rightCountA + func(left, right, s, a, b));
        }

        return currentCost;
    }

    int minimumDeletions(string s) {
        int n = s.size();
        int a = 0, b = 0;

        // Count total 'a's and 'b's
        for (char ch : s) {
            if (ch == 'a') a++;
            else b++;
        }

        return func(0, n - 1, s, a, b);
    }
};

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