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:
    def minAnagramLength(self, s: str) -> int:
        def check(k: int) -> bool:
            for i in range(0, n, k):
                cnt1 = Counter(s[i : i + k])
                for c, v in cnt.items():
                    if cnt1[c] * (n // k) != v:
                        return False
            return True

        cnt = Counter(s)
        n = len(s)
        for i in range(1, n + 1):
            if n % i == 0 and check(i):
                return i
