<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:
        n = len(s)
        # Prefix sums for the number of 1s up to each index
        prefix1 = [0] * (n + 1)
        # Suffix sums for the number of 0s from each index
        suffix0 = [0] * (n + 1)
        
        # Fill prefix1 where prefix1[i] is the number of 1s in s[:i]
        for i in range(1, n + 1):
            prefix1[i] = prefix1[i - 1] + (s[i - 1] == '1')
        
        # Fill suffix0 where suffix0[i] is the number of 0s in s[i:]
        for i in range(n - 1, -1, -1):
            suffix0[i] = suffix0[i + 1] + (s[i] == '0')
        
        # Calculate the minimum flips required
        min_flips = float('inf')
        for i in range(n + 1):
            min_flips = min(min_flips, prefix1[i] + suffix0[i])
        
        return min_flips

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