<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 minimumDeletions(string s) {
        vector<int> cost;

        int bCount = s[0] == 'b';
        bool seenA = false;
        for (int i = 1; i < s.size() - 1; ++i)
        {
            seenA = seenA || s[i] == 'a';

            if (s[i] != s[i + 1] || s[i] != s[i - 1])
            {
                cost.push_back(seenA ? bCount : 0);
            }

            bCount += s[i] == 'b';
        }

        if (cost.size() == 0)
        {
            return 0;
        }

        int aCount = s.back() == 'a';
        bool seenB = false;
        int j = cost.size() - 1;
        int minn = INT_MAX;
        for (int i = s.size() - 2; i >= 1; --i)
        {
            seenB = seenB || s[i] == 'b';

            if (s[i] != s[i-1] || s[i] != s[i+1])
            {
                if (seenB)
                    cost[j] += aCount;
                minn = min(minn, cost[j]);
                j--;
            }

            aCount += s[i] == 'a';
        }

        return minn;
    }
};
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