Minimum Number of Swaps to Make the Binary String Alternating
<p>Given a binary string <code>s</code>, return <em>the <strong>minimum</strong> number of character swaps to make it <strong>alternating</strong>, or </em><code>-1</code><em> if it is impossible.</em></p>

<p>The string is called <strong>alternating</strong> if no two adjacent characters are equal. For example, the strings <code>&quot;010&quot;</code> and <code>&quot;1010&quot;</code> are alternating, while the string <code>&quot;0100&quot;</code> is not.</p>

<p>Any two characters may be swapped, even if they are&nbsp;<strong>not adjacent</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;111000&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Swap positions 1 and 4: &quot;1<u>1</u>10<u>0</u>0&quot; -&gt; &quot;1<u>0</u>10<u>1</u>0&quot;
The string is now alternating.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;010&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> The string is already alternating, no swaps are needed.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;1110&quot;
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
Greedy; String
func minSwaps(s string) int {
	n0 := strings.Count(s, "0")
	n1 := len(s) - n0
	if abs(n0-n1) > 1 {
		return -1
	}
	calc := func(c int) int {
		cnt := 0
		for i, ch := range s {
			x := int(ch - '0')
			if i&1^c != x {
				cnt++
			}
		}
		return cnt / 2
	}
	if n0 == n1 {
		return min(calc(0), calc(1))
	}
	if n0 > n1 {
		return calc(0)
	}
	return calc(1)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}
