# Solving Leetcode Interviews in Seconds with AI: Minimum Length of Anagram Concatenation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3138" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a string s, which is known to be a concatenation of anagrams of some string t. Return the minimum possible length of the string t. An anagram is formed by rearranging the letters of a string. For example, "aab", "aba", and, "baa" are anagrams of "aab".   Example 1:  Input: s = "abba" Output: 2 Explanation: One possible string t could be "ba".  Example 2:  Input: s = "cdef" Output: 4 Explanation: One possible string t could be "cdef", notice that t can be equal to s.  Example 2:  Input: s = "abcbcacabbaccba" Output: 3    Constraints:  1 <= s.length <= 105 s consist only of lowercase English letters.  

	# Explanation
	Here's a solution approach:

*   **Character Frequency Analysis:** Calculate the frequency of each character in the input string `s`. The length of the repeating anagram `t` can be determined by figuring out the smallest number of distinct characters such that their counts divide all the character counts in `s`.
*   **GCD Calculation:** The length of `t` is equal to the number of distinct characters in the original string `t`. If we consider the counts of each character in `s`, then the length of `t` is the greatest common divisor (GCD) of these counts.
*   **Iterate and Validate:** Start with a candidate length `l` (which must divide the GCD calculated earlier). For each potential `l`, form a string `t` using the first `l` distinct characters in sorted order and ensure that `s` can be constructed using anagrams of `t`.

*   **Runtime Complexity:** O(N + K log K), where N is the length of the string s, and K is the number of distinct characters in s (at most 26). Storage Complexity: O(K), mainly for storing character counts.

	
	# Code
	```python
	from math import gcd

def min_anagram_length(s):
    """
    Finds the minimum possible length of the string t, such that s is a concatenation of anagrams of t.

    Args:
        s: The input string.

    Returns:
        The minimum possible length of t.
    """

    char_counts = {}
    for char in s:
        char_counts[char] = char_counts.get(char, 0) + 1

    if not char_counts:
        return 0

    distinct_chars = list(char_counts.keys())
    num_distinct = len(distinct_chars)

    if num_distinct == 0:
        return 0

    g = 0
    for count in char_counts.values():
        g = gcd(g, count)

    min_len = num_distinct

    for length in range(1, num_distinct + 1):
        if g % length == 0:
            min_len = length
            break

    return min_len
	```