<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>

from functools import cache


class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        n = len(s)
        groups = []
        for c in s:
            c = int(c)
            if groups and c == groups[-1][0]:
                char, count = groups.pop()
                groups.append((char, count + 1))
            else:
                groups.append((c, 1))
        groupsSize = len(groups)

        dp = [[float("inf") for _ in range(2)] for _ in range(groupsSize)]

        def recurse(i, prev):
            if i == groupsSize:
                return 0
            if dp[i][prev] != float("inf"):
                return dp[i][prev]
            c, count = groups[i]
            if prev == 1 and c == 1:
                dp[i][prev] = recurse(i + 1, 1)
            elif prev == 1 and c == 0:
                dp[i][prev] = recurse(i + 1, 1) + count
            else:
                dp[i][prev] = min(recurse(i + 1, 1 - c) + count, recurse(i + 1, c))
            return dp[i][prev]
        
        return recurse(0, 0)

        # dp[0][0] = 0 if groups[0][0] == 0 else groups[0][1]
        # dp[0][1] = 0 if groups[0][0] == 1 else groups[0][1]
        # for i in range(1, groupsSize):
        #     c, count = groups[i]
        #     dp[i][1] = dp[i - 1][1] + (0 if c == 1 else count)
        #     dp[i][0] = min(dp[i - 1][1 - c] + count, dp[i - 1][c])
        # print(dp)

        # return min(dp[groupsSize - 1])


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