<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:
    def minFlipsMonoIncr(self, s: str) -> int:
        dp = [[-1 for y in range(len(s))]for x in range(2)]
        def f(cur,prev):
            if cur == len(s):
                return 0
            if dp[prev][cur]!=-1:
                return dp[prev][cur]
            
            ans = 0
            if s[cur] == "0":
                if prev == 0: ##00--> so we can flip the current and not flip the current
                    ans = min(1+f(cur+1,1),0+f(cur+1,0))
                else: ##10--> so we have to flip the current 
                    ans = 1+f(cur+1,1)
            elif s[cur] == "1":
                if prev == 0: ##01--> so we can flip the current and not flip the current
                    ans = min(1+f(cur+1,0),0+f(cur+1,1))
                else: ##11--> so we don't have to flip
                    ans = f(cur+1,1)
            dp[prev][cur] = ans
            return dp[prev][cur]
        return f(0,0)




        
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