<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 ans(int i,int num,string &s,vector<vector<int>> &dp){
        if(i>=s.size())return 0;
        if(dp[num][i]!=-1)return dp[num][i];

        int ans1 = 1e6;
        if(num==1 && s[i]=='0'){
            ans1 = min(ans1,1 + ans(i+1,1,s,dp));
        }
        else if(num==0 && s[i]=='0'){
            ans1 = min(ans1,ans(i+1,0,s,dp));
            ans1 = min(ans1,1 + ans(i+1,1,s,dp));
        }
        else if(num==0 && s[i]=='1'){
            ans1 = min(ans1, ans(i+1,1,s,dp));
            ans1 = min(ans1,1 + ans(i+1,0,s,dp));
        }
        else if(num==1 && s[i]=='1'){
             ans1 = ans(i+1,1,s,dp);
        }
       return dp[num][i] = ans1;
    }
    int minFlipsMonoIncr(string s) {
        int n = s.size();
        vector<vector<int>> dp(2,vector<int> (n+1,-1));
        return ans(0,0,s,dp); 
    }
};
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