Odd String Difference
<p>You are given an array of equal-length strings <code>words</code>. Assume that the length of each string is <code>n</code>.</p>

<p>Each string <code>words[i]</code> can be converted into a <strong>difference integer array</strong> <code>difference[i]</code> of length <code>n - 1</code> where <code>difference[i][j] = words[i][j+1] - words[i][j]</code> where <code>0 &lt;= j &lt;= n - 2</code>. Note that the difference between two letters is the difference between their <strong>positions</strong> in the alphabet i.e.&nbsp;the position of <code>&#39;a&#39;</code> is <code>0</code>, <code>&#39;b&#39;</code> is <code>1</code>, and <code>&#39;z&#39;</code> is <code>25</code>.</p>

<ul>
	<li>For example, for the string <code>&quot;acb&quot;</code>, the difference integer array is <code>[2 - 0, 1 - 2] = [2, -1]</code>.</li>
</ul>

<p>All the strings in words have the same difference integer array, <strong>except one</strong>. You should find that string.</p>

<p>Return<em> the string in </em><code>words</code><em> that has different <strong>difference integer array</strong>.</em></p>

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

<pre>
<strong>Input:</strong> words = [&quot;adc&quot;,&quot;wzy&quot;,&quot;abc&quot;]
<strong>Output:</strong> &quot;abc&quot;
<strong>Explanation:</strong> 
- The difference integer array of &quot;adc&quot; is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of &quot;wzy&quot; is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of &quot;abc&quot; is [1 - 0, 2 - 1] = [1, 1]. 
The odd array out is [1, 1], so we return the corresponding string, &quot;abc&quot;.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;aaa&quot;,&quot;bob&quot;,&quot;ccc&quot;,&quot;ddd&quot;]
<strong>Output:</strong> &quot;bob&quot;
<strong>Explanation:</strong> All the integer arrays are [0, 0] except for &quot;bob&quot;, which corresponds to [13, -13].
</pre>

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

<ul>
	<li><code>3 &lt;= words.length &lt;= 100</code></li>
	<li><code>n == words[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 20</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>
Array; Hash Table; String
class Solution {
    public String oddString(String[] words) {
        var d = new HashMap<String, List<String>>();
        for (var s : words) {
            int m = s.length();
            var cs = new char[m - 1];
            for (int i = 0; i < m - 1; ++i) {
                cs[i] = (char) (s.charAt(i + 1) - s.charAt(i));
            }
            var t = String.valueOf(cs);
            d.putIfAbsent(t, new ArrayList<>());
            d.get(t).add(s);
        }
        for (var ss : d.values()) {
            if (ss.size() == 1) {
                return ss.get(0);
            }
        }
        return "";
    }
}