<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:
        self.dp = {}
        cnt_0 = 0
        for ch in s:
            if ch == '0':
                cnt_0 += 1
        return self.dfs(s, 0, cnt_0)

    def dfs(self, s, i, cnt_0):
        if i == len(s):
            return 0

        if (i, cnt_0) in self.dp:
            return self.dp[(i, cnt_0)]

        if s[i] == '1':
            r1 = 1 + self.dfs(s, i + 1, cnt_0)
            result = min(r1, cnt_0)
        else:
            result = self.dfs(s, i + 1, cnt_0 - 1)

        self.dp[(i, cnt_0)] = result
        return result

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