<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 t[100001][2];
    int solve(string& s,int n,int idx,int prev){
        if(idx>=n){
            return 0;
        }
        if(t[idx][prev]!=-1){
            return t[idx][prev];
        }
        int flip = INT_MAX;
        int noflip = INT_MAX;
        if(s[idx]=='0'){
            if(prev==0){
                flip = 1 + solve(s,n,idx+1,1);
                noflip = solve(s,n,idx+1,0);
            }else{
                flip = 1 + solve(s,n,idx+1,1);
            }
        }else{
            if(prev==0){
                flip = 1 + solve(s,n,idx+1,0);
                noflip = solve(s,n,idx+1,1);
            }else{
                noflip = solve(s,n,idx+1,1);
            }
        }
        return t[idx][prev] = min(flip,noflip);
    }

    int minFlipsMonoIncr(string s) {
        memset(t,-1,sizeof(t));
        return solve(s,s.length(),0,0);
    }
};
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