<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 dfs(int ind,vector<pair<int,int>> &nums,vector<int> &dp)
    {
        if(ind==nums.size())
            return 0;
        if(dp[ind]!=-1)
            return dp[ind];
        return dp[ind] = min(nums[ind].second,nums[ind].first+dfs(ind+1,nums,dp));
    }
    int minimumDeletions(string s) 
    {
        int n=s.length(),cnt=0,tot_a=-1;
        for(char c:s)
        {
            if(c=='b' && tot_a==-1)
                tot_a=0;
            if(c=='a' && tot_a!=-1)
                tot_a++;
        }
        // cout<<tot_a<<endl;
        vector<pair<int,int>> nums;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='b')
            {
                int j=i,cnt=0,cnt1=0;
                while(j<n && s[j]=='b')
                    cnt++,j++;
                while(j<n && s[j]=='a')
                    cnt1++,j++;
                nums.push_back({cnt,tot_a});
                tot_a-=cnt1;
                i=j-1;
            }
        }
        // for(auto it:nums)
        //     cout<<it.first<<" "<<it.second<<endl;
        int m=nums.size();
        vector<int> dp(m,-1);
        return dfs(0,nums,dp);             
    }
};
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