<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)  # Length of the string
        dp0 = [0] * n  # dp0[i] will hold the minimum flips to have all '0's up to index i
        dp1 = [0] * n  # dp1[i] will hold the minimum flips to have all '1's up to index i
        
        # Initialize the first character cases
        dp0[0] = 0 if s[0] == '0' else 1  # If first character is '0', no flip needed for dp0; otherwise, one flip needed
        dp1[0] = 0 if s[0] == '1' else 1  # If first character is '1', no flip needed for dp1; otherwise, one flip needed
        
        # Iterate through the string starting from the second character
        for i in range(1, n):
            # To maintain all '0's up to i, flip current '1' to '0' if necessary
            dp0[i] = dp0[i-1] + (1 if s[i] == '1' else 0)
            # To maintain all '1's up to i, take the min flips from the previous index (either all '0's or '1's) and flip current '0' to '1' if necessary
            dp1[i] = min(dp0[i-1], dp1[i-1]) + (1 if s[i] == '0' else 0)
        
        # The minimum flips required to make the entire string monotone increasing
        return min(dp0[-1], dp1[-1])
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