Check Distances Between Same Letters
<p>You are given a <strong>0-indexed</strong> string <code>s</code> consisting of only lowercase English letters, where each letter in <code>s</code> appears <strong>exactly</strong> <strong>twice</strong>. You are also given a <strong>0-indexed</strong> integer array <code>distance</code> of length <code>26</code>.</p>

<p>Each letter in the alphabet is numbered from <code>0</code> to <code>25</code> (i.e. <code>&#39;a&#39; -&gt; 0</code>, <code>&#39;b&#39; -&gt; 1</code>, <code>&#39;c&#39; -&gt; 2</code>, ... , <code>&#39;z&#39; -&gt; 25</code>).</p>

<p>In a <strong>well-spaced</strong> string, the number of letters between the two occurrences of the <code>i<sup>th</sup></code> letter is <code>distance[i]</code>. If the <code>i<sup>th</sup></code> letter does not appear in <code>s</code>, then <code>distance[i]</code> can be <strong>ignored</strong>.</p>

<p>Return <code>true</code><em> if </em><code>s</code><em> is a <strong>well-spaced</strong> string, otherwise return </em><code>false</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abaccb&quot;, distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
<strong>Output:</strong> true
<strong>Explanation:</strong>
- &#39;a&#39; appears at indices 0 and 2 so it satisfies distance[0] = 1.
- &#39;b&#39; appears at indices 1 and 5 so it satisfies distance[1] = 3.
- &#39;c&#39; appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since &#39;d&#39; does not appear in s, it can be ignored.
Return true because s is a well-spaced string.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
<strong>Output:</strong> false
<strong>Explanation:</strong>
- &#39;a&#39; appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.
</pre>

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

<ul>
	<li><code>2 &lt;= s.length &lt;= 52</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
	<li>Each letter appears in <code>s</code> exactly twice.</li>
	<li><code>distance.length == 26</code></li>
	<li><code>0 &lt;= distance[i] &lt;= 50</code></li>
</ul>
Array; Hash Table; String
class Solution {
public:
    bool checkDistances(string s, vector<int>& distance) {
        int d[26]{};
        for (int i = 1; i <= s.size(); ++i) {
            int j = s[i - 1] - 'a';
            if (d[j] && i - d[j] - 1 != distance[j]) {
                return false;
            }
            d[j] = i;
        }
        return true;
    }
};