<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.size();
        vector<pair<int,int>> pre(n,{0,0});
        vector<pair<int,int>> suff(n,{0,0});
        for(int i=0;i<s.size();i++){
            if(i==0){
                if(s[i]=='0'){
                    pre[i]={1,0};
                }
                else{
                    pre[i]={0,1};
                }
            }
            else if(s[i]=='0'){
                pre[i].first=pre[i-1].first+1;
                pre[i].second=pre[i-1].second;
            }
            else if(s[i]=='1'){
                pre[i].first=pre[i-1].first;
                pre[i].second=pre[i-1].second+1;
            }
        }
        for(int i=s.size()-1;i>=0;i--){
            if(i==s.size()-1){
                if(s[i]=='0'){
                    suff[i]={1,0};
                }
                else{
                    suff[i]={0,1};
                }
            }
            else if(s[i]=='0'){
                suff[i].first=suff[i+1].first+1;
                suff[i].second=suff[i+1].second;
            }
            else if(s[i]=='1'){
                suff[i].first=suff[i+1].first;
                suff[i].second=suff[i+1].second+1; 
            }
        }
        int ans=INT_MAX;
        for(int i=0;i<s.size();i++){
            if(i==0){
                ans=min(ans,suff[i+1].first);
            }
            else if(i==s.size()-1){
                ans=min(ans,pre[i-1].second);
            }
            else{
                int total=pre[i-1].second+suff[i+1].first;
                ans=min(ans,total);
            }
        }
        return ans;
    }
};
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