<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 count_a=0;
        int count_b=0;
        int count=0;
        vector<int>arr1;
        vector<int>arr2;
        int min=0;
        int i=0;
        int j=s.size()-1;
        while(i<s.size()&&s[i]=='a')i++;
        while(j>0&&s[j]=='b')j--;
        while(i<=j){
           
            while(i<=j&&s[i]=='b'){
                count_b++;
                i++;
            }
            arr1.push_back(count_b);
             while(i<=j&&s[i]=='a'){
                count_a++;
                i++;
            }
           arr2.push_back(count_a);
           count_a=0;
        }
        int sum=0;
        if(arr1.size()>0&&arr2.size()>0){
            for(i=arr2.size()-1;i>=0;i--){
            sum+=arr2[i];
            arr2[i]=sum;
        }
        min=arr2[0];
        j=1;
        for(int i=0;i<arr1.size();i++){
            if(j<arr2.size()&&min>arr1[i]+arr2[j])min=arr1[i]+arr2[j];
            j++;
        }
        if(min>arr1[arr1.size()-1])min=arr1[arr1.size()-1];
        }

        
        return min;
    }
};
44170