<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(int i, int prev, string s, vector<vector<int>>&mem){
    if(i>=s.length()){
       return mem[i][prev]= 0;
    }
    if(mem[i][prev] != -1){return mem[i][prev];}
    int ans = INT_MAX;
    if(prev==1){
        if(s[i]=='1'){
            ans = min(ans, solve(i+1, 1, s, mem));
        }
        else{
            ans = min(ans, 1 + solve(i+1, 1, s, mem));
        }
    }
    else{
        if(s[i]=='0'){
            ans = min(ans, solve( i+1, 0, s, mem));
            ans = min(ans, 1+solve(i+1, 1, s, mem));
        }
        else{
            ans = min(ans, solve(i+1, 1, s, mem));
            ans = min(ans, 1+solve(i+1, 0, s, mem));
        }
    }
    return mem[i][prev]=ans;
}
    int minFlipsMonoIncr(string s) {
        // vector<vector<int>>mem(s.length()+1, vector<int>(2, -1));
        // return solve(0, 0, s, mem);
    int n = s.length();
        vector<vector<int>>dp(s.length()+1, vector<int>(2));
        for(int i=0; i<2; i++){
            dp[n][1]=0;
        }
        for(int i=n-1; i>=0; i--){
            for(int prev=0; prev<2; prev++){
                int ans = INT_MAX;
                if(prev==1){
                    if(s[i]=='1'){
                        ans = min(ans, dp[i+1][1]);
                    }
                    else{
                        ans = min(ans, 1 + dp[i+1][1]);
                    }
                }
                else{
                    if(s[i]=='0'){
                        ans = min(ans, dp[i+1][0]);
                        ans = min(ans, 1+dp[i+1][1]);
                    }
                    else{
                        ans = min(ans, dp[i+1][1]);
                        ans = min(ans, 1+dp[i+1][0]);
                    }
                }
                dp[i][prev]=ans;
            }
        }
        return dp[0][0];

        
    }
};
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