rearrange-string-to-avoid-character-pair

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Algorithms

You are given a string <code>s</code> and two distinct lowercase English letters <code>x</code> and <code>y</code>.

Rearrange the characters of <code>s</code> to construct a new string <code>t</code> such that:

<ul>
	<li><code>t</code> is a permutation of <code>s</code>.</li>
	<li>Every occurrence of <code>y</code> appears before every occurrence of <code>x</code> in <code>t</code>.</li>
</ul>

Return any valid string <code>t</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: s = &quot;aabc&quot;, x = &quot;a&quot;, y = &quot;c&quot;

Output: &quot;cbaa&quot;

Explanation:

The string <code>&quot;cbaa&quot;</code> is a permutation of <code>&quot;aabc&quot;</code>, and every occurrence of <code>&#39;c&#39;</code> appears before every occurrence of <code>&#39;a&#39;</code>.
</div>

Example 2:

<div class="example-block">
Input: s = &quot;dcab&quot;, x = &quot;d&quot;, y = &quot;b&quot;

Output: &quot;cabd&quot;

Explanation:

The string <code>&quot;cabd&quot;</code> is a permutation of <code>&quot;dcab&quot;</code>, and every occurrence of <code>&#39;b&#39;</code> appears before every occurrence of <code>&#39;d&#39;</code>.
</div>

Example 3:

<div class="example-block">
Input: s = &quot;axe&quot;, x = &quot;o&quot;, y = &quot;x&quot;

Output: &quot;axe&quot;

Explanation:

The string <code>&quot;axe&quot;</code> is already valid. Since <code>&#39;o&#39;</code> does not occur in the string, the required condition is automatically satisfied.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
	<li><code>x</code> and <code>y</code> are lowercase English letters.</li>
	<li><code>x != y</code></li>
</ul>