Minimum Swaps to Make Strings Equal
<p>You are given two strings <code>s1</code> and <code>s2</code> of equal length consisting of letters <code>&quot;x&quot;</code> and <code>&quot;y&quot;</code> <strong>only</strong>. Your task is to make these two strings equal to each other. You can swap any two characters that belong to <strong>different</strong> strings, which means: swap <code>s1[i]</code> and <code>s2[j]</code>.</p>

<p>Return the minimum number of swaps required to make <code>s1</code> and <code>s2</code> equal, or return <code>-1</code> if it is impossible to do so.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;xx&quot;, s2 = &quot;yy&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Swap s1[0] and s2[1], s1 = &quot;yx&quot;, s2 = &quot;yx&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;xy&quot;, s2 = &quot;yx&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Swap s1[0] and s2[0], s1 = &quot;yy&quot;, s2 = &quot;xx&quot;.
Swap s1[0] and s2[1], s1 = &quot;xy&quot;, s2 = &quot;xy&quot;.
Note that you cannot swap s1[0] and s1[1] to make s1 equal to &quot;yx&quot;, cause we can only swap chars in different strings.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;xx&quot;, s2 = &quot;xy&quot;
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 1000</code></li>
	<li><code>s1.length == s2.length</code></li>
	<li><code>s1, s2</code> only contain <code>&#39;x&#39;</code> or <code>&#39;y&#39;</code>.</li>
</ul>
Greedy; Math; String
function minimumSwap(s1: string, s2: string): number {
    let xy = 0,
        yx = 0;

    for (let i = 0; i < s1.length; ++i) {
        const a = s1[i],
            b = s2[i];
        xy += a < b ? 1 : 0;
        yx += a > b ? 1 : 0;
    }

    if ((xy + yx) % 2 !== 0) {
        return -1;
    }

    return Math.floor(xy / 2) + Math.floor(yx / 2) + (xy % 2) + (yx % 2);
}
