<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) {
        
        
        int n=s.length();
        vector<int> left0(n),left1(n),right0(n),right1(n);
        int cnt0=0,cnt1=0;
        for(int i=0;i<n;i++){
            if(s[i]=='0') cnt0++;
            left0[i]=cnt0;
            if(s[i]=='1') cnt1++;
            left1[i]=cnt1;
        }
        cnt0=0,cnt1=0;
        for(int i=n-1;i>=0;i--){
            if(s[i]=='0') cnt0++;
            right0[i]=cnt0;
            if(s[i]=='1') cnt1++;
            right1[i]=cnt1;
        }
        
        int res=n;
        for(int i=0;i<n-1;i++){
            res=min(res,left1[i]+right0[i+1]);
            res=min(res,left1[i]+right1[i+1]);
            res=min(res,left0[i]+right0[i+1]);
        }
        
        return res;
    }
    
};
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