Minimum Length of Anagram Concatenation
<p>You are given a string <code>s</code>, which is known to be a concatenation of <strong>anagrams</strong> of some string <code>t</code>.</p>

<p>Return the <strong>minimum</strong> possible length of the string <code>t</code>.</p>

<p>An <strong>anagram</strong> is formed by rearranging the letters of a string. For example, &quot;aab&quot;, &quot;aba&quot;, and, &quot;baa&quot; are anagrams of &quot;aab&quot;.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abba&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>One possible string <code>t</code> could be <code>&quot;ba&quot;</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;cdef&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>One possible string <code>t</code> could be <code>&quot;cdef&quot;</code>, notice that <code>t</code> can be equal to <code>s</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abcbcacabbaccba&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consist only of lowercase English letters.</li>
</ul>
Hash Table; String; Counting
class Solution {
public:
    int minAnagramLength(string s) {
        int n = s.size();
        int cnt[26]{};
        for (char c : s) {
            cnt[c - 'a']++;
        }
        auto check = [&](int k) {
            for (int i = 0; i < n; i += k) {
                int cnt1[26]{};
                for (int j = i; j < i + k; ++j) {
                    cnt1[s[j] - 'a']++;
                }
                for (int j = 0; j < 26; ++j) {
                    if (cnt1[j] * (n / k) != cnt[j]) {
                        return false;
                    }
                }
            }
            return true;
        };
        for (int i = 1;; ++i) {
            if (n % i == 0 && check(i)) {
                return i;
            }
        }
    }
};