Replace All Digits with Characters
<p>You are given a <strong>0-indexed</strong> string <code>s</code> that has lowercase English letters in its <strong>even</strong> indices and digits in its <strong>odd</strong> indices.</p>

<p>You must perform an operation <code>shift(c, x)</code>, where <code>c</code> is a character and <code>x</code> is a digit, that returns the <code>x<sup>th</sup></code> character after <code>c</code>.</p>

<ul>
	<li>For example, <code>shift(&#39;a&#39;, 5) = &#39;f&#39;</code> and <code>shift(&#39;x&#39;, 0) = &#39;x&#39;</code>.</li>
</ul>

<p>For every <strong>odd</strong> index <code>i</code>, you want to replace the digit <code>s[i]</code> with the result of the <code>shift(s[i-1], s[i])</code> operation.</p>

<p>Return <code>s</code><em> </em>after replacing all digits. It is <strong>guaranteed</strong> that<em> </em><code>shift(s[i-1], s[i])</code><em> </em>will never exceed<em> </em><code>&#39;z&#39;</code>.</p>

<p><strong>Note</strong> that <code>shift(c, x)</code> is <strong>not</strong> a preloaded function, but an operation <em>to be implemented</em> as part of the solution.</p>

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

<pre>
<strong>Input:</strong> s = &quot;a1c1e1&quot;
<strong>Output:</strong> &quot;abcdef&quot;
<strong>Explanation: </strong>The digits are replaced as follows:
- s[1] -&gt; shift(&#39;a&#39;,1) = &#39;b&#39;
- s[3] -&gt; shift(&#39;c&#39;,1) = &#39;d&#39;
- s[5] -&gt; shift(&#39;e&#39;,1) = &#39;f&#39;</pre>

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

<pre>
<strong>Input:</strong> s = &quot;a1b2c3d4e&quot;
<strong>Output:</strong> &quot;abbdcfdhe&quot;
<strong>Explanation: </strong>The digits are replaced as follows:
- s[1] -&gt; shift(&#39;a&#39;,1) = &#39;b&#39;
- s[3] -&gt; shift(&#39;b&#39;,2) = &#39;d&#39;
- s[5] -&gt; shift(&#39;c&#39;,3) = &#39;f&#39;
- s[7] -&gt; shift(&#39;d&#39;,4) = &#39;h&#39;</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists only of lowercase English letters and digits.</li>
	<li><code>shift(s[i-1], s[i]) &lt;= &#39;z&#39;</code> for all <strong>odd</strong> indices <code>i</code>.</li>
</ul>
String
char* replaceDigits(char* s) {
    int n = strlen(s);
    for (int i = 1; i < n; i += 2) {
        s[i] = s[i - 1] + s[i] - '0';
    }
    return s;
}