Minimum Number of Changes to Make Binary String Beautiful
<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> having an even length.</p>

<p>A string is <strong>beautiful</strong> if it&#39;s possible to partition it into one or more substrings such that:</p>

<ul>
	<li>Each substring has an <strong>even length</strong>.</li>
	<li>Each substring contains <strong>only</strong> <code>1</code>&#39;s or <strong>only</strong> <code>0</code>&#39;s.</li>
</ul>

<p>You can change any character in <code>s</code> to <code>0</code> or <code>1</code>.</p>

<p>Return <em>the <strong>minimum</strong> number of changes required to make the string </em><code>s</code> <em>beautiful</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1001&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We change s[1] to 1 and s[3] to 0 to get string &quot;1100&quot;.
It can be seen that the string &quot;1100&quot; is beautiful because we can partition it into &quot;11|00&quot;.
It can be proven that 2 is the minimum number of changes needed to make the string beautiful.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;10&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We change s[1] to 1 to get string &quot;11&quot;.
It can be seen that the string &quot;11&quot; is beautiful because we can partition it into &quot;11&quot;.
It can be proven that 1 is the minimum number of changes needed to make the string beautiful.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> We don&#39;t need to make any changes as the string &quot;0000&quot; is beautiful already.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> has an even length.</li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
String
class Solution:
    def minChanges(self, s: str) -> int:
        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))
