<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 f(int indx, int isZero, string s, vector<vector<int>>&dp){
        if(indx == s.length()) return 0;
        if(dp[indx][isZero] != -1) return dp[indx][isZero];

        int same = 1e9, diff = 1e9;
        if(isZero && s[indx] == '0' || !isZero && s[indx] == '1'){
            same = min(same, f(indx + 1,isZero, s, dp));
        }

        else if(isZero && s[indx] == '1'){//expecting 0, but had 1
            //2 options change this, from now on accept 1
            int change = 1 + f(indx + 1, isZero, s, dp);
            int acceptOne = f(indx + 1, !isZero, s, dp);
            diff = min(change, acceptOne);
        }

        else if(!isZero && s[indx] == '0'){
            diff = min(diff, 1 + f(indx + 1, isZero, s, dp));
        }

        return dp[indx][isZero] = min(same, diff);
    }

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

        // return f(0, 1, s, dp);

        int n = s.length();
        vector<vector<int>>dp(n + 1, vector<int>(2, 0));


        for(int indx = n - 1; indx >= 0; indx --){
            for(int isZero = 0; isZero <= 1; isZero ++){
                int same = 1e9, diff = 1e9;
                if(isZero && s[indx] == '0' || !isZero && s[indx] == '1'){
                    same = min(same, dp[indx + 1][isZero] );
                }

                else if(isZero && s[indx] == '1'){//expecting 0, but had 1
                    //2 options change this, from now on accept 1
                    int change = 1 + dp[indx + 1][isZero];
                    int acceptOne = dp[indx + 1][!isZero];
                    diff = min(change, acceptOne);
                }

                else if(!isZero && s[indx] == '0'){
                    diff = min(diff, 1 + dp[indx + 1][isZero]);
                }

                dp[indx][isZero] = min(same, diff);
            }
        }

        return dp[0][1];
    }
};
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