<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 solve(string &s , int curr_index , int prev, vector<vector<int>> &dp){
        if(curr_index >= s.length()) return 0;

        if(dp[curr_index][prev] != -1) return dp[curr_index][prev];

        int flip = INT_MAX;
        int noFlip = INT_MAX;

        if(s[curr_index] == '0'){
            if(prev == 0){
                flip = 1 + solve(s,curr_index+1, 1, dp);
                noFlip = solve(s, curr_index+1, 0, dp);
            }
            else{
                //prev == '1'
                flip = 1 + solve(s, curr_index+1,1, dp);
             
            }

        }
        else if(s[curr_index] == '1'){
            if(prev == 0){
                flip = 1 + solve(s, curr_index+1,0, dp);
                noFlip = solve(s, curr_index+1,1, dp);


            }
            else{
                //prev == '1'
                noFlip =  solve(s, curr_index+1, 1, dp);
            }
        }
        return dp[curr_index][prev] = min(flip, noFlip);
    }
    

    int minFlipsMonoIncr(string s) {
        int n = s.length();
        vector<vector<int>> dp(n+1, vector<int>(2, -1));
        return solve(s, 0, 0, dp);
       
    }
};
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