Shortest Distance to Target String in a Circular Array
<p>You are given a <strong>0-indexed</strong> <strong>circular</strong> string array <code>words</code> and a string <code>target</code>. A <strong>circular array</strong> means that the array&#39;s end connects to the array&#39;s beginning.</p>

<ul>
	<li>Formally, the next element of <code>words[i]</code> is <code>words[(i + 1) % n]</code> and the previous element of <code>words[i]</code> is <code>words[(i - 1 + n) % n]</code>, where <code>n</code> is the length of <code>words</code>.</li>
</ul>

<p>Starting from <code>startIndex</code>, you can move to either the next word or the previous word with <code>1</code> step at a time.</p>

<p>Return <em>the <strong>shortest</strong> distance needed to reach the string</em> <code>target</code>. If the string <code>target</code> does not exist in <code>words</code>, return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;hello&quot;,&quot;i&quot;,&quot;am&quot;,&quot;leetcode&quot;,&quot;hello&quot;], target = &quot;hello&quot;, startIndex = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> We start from index 1 and can reach &quot;hello&quot; by
- moving 3 units to the right to reach index 4.
- moving 2 units to the left to reach index 4.
- moving 4 units to the right to reach index 0.
- moving 1 unit to the left to reach index 0.
The shortest distance to reach &quot;hello&quot; is 1.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;b&quot;,&quot;leetcode&quot;], target = &quot;leetcode&quot;, startIndex = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> We start from index 0 and can reach &quot;leetcode&quot; by
- moving 2 units to the right to reach index 3.
- moving 1 unit to the left to reach index 3.
The shortest distance to reach &quot;leetcode&quot; is 1.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;i&quot;,&quot;eat&quot;,&quot;leetcode&quot;], target = &quot;ate&quot;, startIndex = 0
<strong>Output:</strong> -1
<strong>Explanation:</strong> Since &quot;ate&quot; does not exist in <code>words</code>, we return -1.
</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
	<li><code>words[i]</code> and <code>target</code> consist of only lowercase English letters.</li>
	<li><code>0 &lt;= startIndex &lt; words.length</code></li>
</ul>
Array; String
impl Solution {
    pub fn closet_target(words: Vec<String>, target: String, start_index: i32) -> i32 {
        let start_index = start_index as usize;
        let n = words.len();
        for i in 0..=n >> 1 {
            if words[(start_index - i + n) % n] == target || words[(start_index + i) % n] == target
            {
                return i as i32;
            }
        }
        -1
    }
}
