<p>A binary string is monotone increasing if it consists of some number of <code>0</code>&#39;s (possibly none), followed by some number of <code>1</code>&#39;s (also possibly none).</p>

<p>You are given a binary string <code>s</code>. You can flip <code>s[i]</code> changing it from <code>0</code> to <code>1</code> or from <code>1</code> to <code>0</code>.</p>

<p>Return <em>the minimum number of flips to make </em><code>s</code><em> monotone increasing</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00110&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We flip the last digit to get 00111.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;010110&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We flip to get 011111, or alternatively 000111.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00011000&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We flip to get 00000000.
</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 either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int minFlipsMonoIncr(string s) {
       vector<int> ze,on;
       
       int cnt=0,n=s.size();
        int ans=n;
       for(int i=0;i<n;i++)
       { 
          if(s[i]=='0')
          cnt++;
          ze.push_back(cnt);
       } 
       ans=min(ans,cnt);
      
       cnt=0;
       for(int i=n-1;i>=0;i--)
       {
         if(s[i]=='1')
         cnt++;
         on.push_back(cnt);
       }
       ans=min(ans,cnt);
       reverse(on.begin(),on.end());
       
      
       for(int i=n-1;i>=0;i--)
       {
           int an=0;
           if(s[i]=='0')
           {
              an=n+1-ze[i]-on[i];
           }
           else
           {
             
             an=n-ze[i]-on[i];
           }
           ans=min(an,ans);
       }
       return ans;
    }
};
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