subsequence-after-one-replacement

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Algorithms

You are given two strings <code>s</code> and <code>t</code> consisting of lowercase English letters.

You may choose at most one index in <code>s</code> and replace the character at that index with any lowercase English letter.

Return <code>true</code> if it is possible to make <code>s</code> a subsequence of <code>t</code>; otherwise, return <code>false</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: s = &quot;cat&quot;, t = &quot;chat&quot;

Output: true

Explanation:

<ul>
	<li>Replace <code>s[1]</code> from <code>&#39;a&#39;</code> to <code>&#39;h&#39;</code>. The resulting string is <code>&quot;cht&quot;</code>.</li>
	<li><code>&quot;cht&quot;</code> is a subsequence of <code>&quot;chat&quot;</code> because we can match <code>&#39;c&#39;</code>, <code>&#39;h&#39;</code>, and <code>&#39;t&#39;</code> in order.</li>
</ul>
</div>

Example 2:

<div class="example-block">
Input: s = &quot;plane&quot;, t = &quot;apple&quot;

Output: false

Explanation:

<ul>
	<li>The characters <code>&#39;p&#39;</code>, <code>&#39;l&#39;</code>, and <code>&#39;e&#39;</code> can be matched in <code>t</code>, but the remaining characters cannot be matched while preserving the required order.</li>
	<li>Even after replacing any one character in <code>s</code>, it is impossible to make <code>s</code> a subsequence of <code>t</code>.</li>
</ul>
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length, t.length &lt;= 105</code></li>
	<li><code>s</code> and <code>t</code> consist only of lowercase English letters.</li>
</ul>