<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<int> pre(n,0),suff(n,0);
        pre[0] = s[0] == '1' ? 1 : 0;
        suff[n-1] = s[n-1] == '0' ? 1 : 0;
        for(int i = 1; i < n;i++){
            pre[i] = s[i] == '1' ? 1 + pre[i-1] : pre[i-1];
        }
        for(int j = n - 2; j >= 0;j--){
            suff[j] = s[j] == '0' ? 1 + suff[j+1] : suff[j+1];
        }
        int ans = INT_MAX;
        for(int i = 0; i < n;i++) ans = min(ans,pre[i] + suff[i] - 1);
        return ans;
    }
};
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