<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:
    vector<vector<int>>dp;
    int rec(int index, int isOne, string& s) {
        if (index == s.length()) {
            return 0;
        }
        if(dp[index][isOne] != -1) return dp[index][isOne];
        int ans = 1e9;
        if (isOne == 1) {
            if (s[index] == '1') {
                ans = min(ans, rec(index + 1, 1, s)); // skip the flip
            } else {
                ans = min(ans, 1 + rec(index + 1, 1, s)); // flip the bit
            }

        } else {
            if (s[index] == '1') {
                ans = min(ans, rec(index + 1, 1, s));     // skip the flip
                ans = min(ans, 1 + rec(index + 1, 0, s)); // flip the bit
            } else {
                ans = min(ans, rec(index + 1, 0, s));     // skip the flip
                ans = min(ans, 1 + rec(index + 1, 1, s)); // flip the bit
            }
        }

        return dp[index][isOne] = ans;
    }
    int minFlipsMonoIncr(string s) {
        dp.assign(s.length() + 1, vector<int>(2, -1));
        return rec(0, 0, s); 
    }
};
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