<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:
        # cache={}
        # def flips(i,pre):
        #     if i==len(s):
        #         return 0
        #     if (i,pre) in cache:
        #         return cache[(i,pre)]
        #     if s[i]=='0':
        #         if pre=='0':
        #             res=min(flips(i+1,'0'),1+flips(i+1,'1'))
        #         else:
        #             res=1+flips(i+1,'1')
        #     else:
        #         if pre=='1':
        #             res=flips(i+1,'1')
        #         else:
        #             res=min(1+flips(i+1,'0'),flips(i+1,'1'))
        #     cache[(i,pre)]=res
        #     return res
        # return flips(0,'0')

        cache=[[-1]*2 for _ in range(len(s))]
        def flips(i,pre):
            if i==len(s):
                return 0
            if cache[i][pre]!=-1:
                return cache[i][pre]
            if s[i]=='0':
                if pre==0:
                    res=min(flips(i+1,0),1+flips(i+1,1))
                else:
                    res=1+flips(i+1,1)
            else:
                if pre==1:
                    res=flips(i+1,1)
                else:
                    res=min(1+flips(i+1,0),flips(i+1,1))
            cache[i][pre]=res
            return res
        return flips(0,0)
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