Replace All 's to Avoid Consecutive Repeating Characters
<p>Given a string <code>s</code> containing only lowercase English letters and the <code>&#39;?&#39;</code> character, convert <strong>all </strong>the <code>&#39;?&#39;</code> characters into lowercase letters such that the final string does not contain any <strong>consecutive repeating </strong>characters. You <strong>cannot </strong>modify the non <code>&#39;?&#39;</code> characters.</p>

<p>It is <strong>guaranteed </strong>that there are no consecutive repeating characters in the given string <strong>except </strong>for <code>&#39;?&#39;</code>.</p>

<p>Return <em>the final string after all the conversions (possibly zero) have been made</em>. If there is more than one solution, return <strong>any of them</strong>. It can be shown that an answer is always possible with the given constraints.</p>

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

<pre>
<strong>Input:</strong> s = &quot;?zs&quot;
<strong>Output:</strong> &quot;azs&quot;
<strong>Explanation:</strong> There are 25 solutions for this problem. From &quot;azs&quot; to &quot;yzs&quot;, all are valid. Only &quot;z&quot; is an invalid modification as the string will consist of consecutive repeating characters in &quot;zzs&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ubv?w&quot;
<strong>Output:</strong> &quot;ubvaw&quot;
<strong>Explanation:</strong> There are 24 solutions for this problem. Only &quot;v&quot; and &quot;w&quot; are invalid modifications as the strings will consist of consecutive repeating characters in &quot;ubvvw&quot; and &quot;ubvww&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consist of lowercase English letters and <code>&#39;?&#39;</code>.</li>
</ul>
String
func modifyString(s string) string {
	n := len(s)
	cs := []byte(s)
	for i := range s {
		if cs[i] == '?' {
			for c := byte('a'); c <= byte('c'); c++ {
				if (i > 0 && cs[i-1] == c) || (i+1 < n && cs[i+1] == c) {
					continue
				}
				cs[i] = c
				break
			}
		}
	}
	return string(cs)
}