Shortest Distance to a Character
<p>Given a string <code>s</code> and a character <code>c</code> that occurs in <code>s</code>, return <em>an array of integers </em><code>answer</code><em> where </em><code>answer.length == s.length</code><em> and </em><code>answer[i]</code><em> is the <strong>distance</strong> from index </em><code>i</code><em> to the <strong>closest</strong> occurrence of character </em><code>c</code><em> in </em><code>s</code>.</p>

<p>The <strong>distance</strong> between two indices <code>i</code> and <code>j</code> is <code>abs(i - j)</code>, where <code>abs</code> is the absolute value function.</p>

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

<pre>
<strong>Input:</strong> s = &quot;loveleetcode&quot;, c = &quot;e&quot;
<strong>Output:</strong> [3,2,1,0,1,0,0,1,2,2,1,0]
<strong>Explanation:</strong> The character &#39;e&#39; appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of &#39;e&#39; for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of &#39;e&#39; for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the &#39;e&#39; at index 3 and the &#39;e&#39; at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of &#39;e&#39; for index 8 is at index 6, so the distance is abs(8 - 6) = 2.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaab&quot;, c = &quot;b&quot;
<strong>Output:</strong> [3,2,1,0]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s[i]</code> and <code>c</code> are lowercase English letters.</li>
	<li>It is guaranteed that <code>c</code> occurs at least once in <code>s</code>.</li>
</ul>
Array; Two Pointers; String
func shortestToChar(s string, c byte) []int {
	n := len(s)
	ans := make([]int, n)
	const inf int = 1 << 30
	pre := -inf
	for i := range s {
		if s[i] == c {
			pre = i
		}
		ans[i] = i - pre
	}
	suf := inf
	for i := n - 1; i >= 0; i-- {
		if s[i] == c {
			suf = i
		}
		ans[i] = min(ans[i], suf-i)
	}
	return ans
}