Minimum Changes To Make Alternating Binary String
<p>You are given a string <code>s</code> consisting only of the characters <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>. In one operation, you can change any <code>&#39;0&#39;</code> to <code>&#39;1&#39;</code> or vice versa.</p>

<p>The string is called alternating if no two adjacent characters are equal. For example, the string <code>&quot;010&quot;</code> is alternating, while the string <code>&quot;0100&quot;</code> is not.</p>

<p>Return <em>the <strong>minimum</strong> number of operations needed to make</em> <code>s</code> <em>alternating</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;0100&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> If you change the last character to &#39;1&#39;, s will be &quot;0101&quot;, which is alternating.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;10&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> s is already alternating.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;1111&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need two operations to reach &quot;0101&quot; or &quot;1010&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
String
func minOperations(s string) int {
	cnt := 0
	for i, c := range s {
		if c != []rune("01")[i&1] {
			cnt++
		}
	}
	return min(cnt, len(s)-cnt)
}