<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:
        memo = {}

        def helper(i, prev):
            # Base case: if we've processed the entire string
            if i == len(s):
                return 0

            # Memoization check
            if (i, prev) in memo:
                return memo[(i, prev)]

            # Current character is '1'
            if s[i] == '1':
                if prev == 0:
                    # Option 1: Keep it as '1', Option 2: Flip it to '0'
                    memo[(i, prev)] = min(helper(i + 1, 1), 1 + helper(i + 1, 0))
                else:
                    # We keep it as '1' since prev is already 1
                    memo[(i, prev)] = helper(i + 1, 1)
            else:  # Current character is '0'
                if prev == 1:
                    # Must flip to '1' because previous character is '1'
                    memo[(i, prev)] = 1 + helper(i + 1, 1)
                else:
                    # Option 1: Keep it as '0', Option 2: Flip to '1'
                    memo[(i, prev)] = min(helper(i + 1, 0), 1 + helper(i + 1, 1))

            return memo[(i, prev)]

        # Start the recursion
        result = helper(0, 0)
   
        return result

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