<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) {
        int n = s.length(), i = 0, j = n-1;
        while(i<n && s[i] == 'a') i++;
        while(j>=0 && s[j] == 'b') j--;
        vector<vector<int>> arr(2);
        int count = 0, prev = '.', counta = 0;
        for(int k = i; k<=j; k++){
            if(s[k] != prev){
                if(count != 0) arr[prev-'a'].push_back(count);
                if(prev == 'a') counta += count;
                count = 1;
                prev = s[k];
            }else count++;
        }
        if(count != 0) arr[prev-'a'].push_back(count);
        if(prev == 'a') counta += count;
        int totala = 0, totalb = 0, ans = counta;
        for(int k = 0; k<arr[0].size(); k++){
            totala += arr[0][k];
            totalb += arr[1][k];
            ans = min(ans, totalb + counta - totala);
        }
        return ans;
    }
};
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