<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 n;

    int solve(string &s,int curr_index,int prev_value,vector<vector<int>> &dp){

        if(curr_index >= n){

            return 0;
        }

        if(dp[curr_index][prev_value] != -1){

            return dp[curr_index][prev_value];
        }

        int flip = INT_MAX;
        int no_flip = INT_MAX;

        if(s[curr_index] == '0'){

            if(prev_value == 0){

                flip = 1 + solve(s,curr_index+1,1,dp);

                no_flip = solve(s,curr_index+1,0,dp);
            }
            else{

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

            if(prev_value == 0){

                flip = 1 + solve(s,curr_index+1,0,dp);

                no_flip = solve(s,curr_index+1,1,dp);
            }
            else{

                no_flip = solve(s,curr_index+1,1,dp);
            }
        }

        return dp[curr_index][prev_value] = min(flip,no_flip);
    }

    int minFlipsMonoIncr(string s) {
        
        n = s.size();

        int curr_index = 0;
        int prev_value = 0;  // taking as zero

        vector<vector<int>> dp(n,vector<int>(2,-1));

        return solve(s,curr_index,prev_value,dp);

    }
};


/*


class Solution {
public:

    int n;

    int solve(string &s,int curr_index,int prev_value){

        if(curr_index >= n){

            return 0;
        }

        int flip = INT_MAX;
        int no_flip = INT_MAX;

        if(s[curr_index] == '0'){

            if(prev_value == 0){

                flip = 1 + solve(s,curr_index+1,1);

                no_flip = solve(s,curr_index+1,0);
            }
            else{

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

            if(prev_value == 0){

                flip = 1 + solve(s,curr_index+1,0);

                no_flip = solve(s,curr_index+1,1);
            }
            else{

                no_flip = solve(s,curr_index+1,1);
            }
        }

        return min(flip,no_flip);
    }

    int minFlipsMonoIncr(string s) {
        
        n = s.size();

        int curr_index = 0;
        int prev_value = 0;  // taking as zero

        return solve(s,curr_index,prev_value);

    }
};


*/
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