#3138

Medium

LeetCode Problem Workspace

Minimum Length of Anagram Concatenation

The problem asks to find the minimum length of a string t such that s is a concatenation of anagrams of t.

Problem Statement

You are given a string s, which is known to be a concatenation of anagrams of some string t. An anagram is formed by rearranging the letters of a string. For example, 'aab', 'aba', and 'baa' are all anagrams of 'aab'. The goal is to return the minimum possible length of the string t.

You need to find the smallest string t such that s can be split into an integer number of anagrams of t. The length of t must divide the length of s. This problem can be efficiently tackled using hash tables to keep track of character counts.

Examples

Example 1

Input: s = "abba"

Output: 2

One possible string t could be "ba" .

Example 2

Input: s = "cdef"

Output: 4

One possible string t could be "cdef" , notice that t can be equal to s .

Example 3

Input: s = "abcbcacabbaccba"

Output: 3

Example details omitted.

Constraints

1 <= s.length <= 105
s consist only of lowercase English letters.

Solution Approach

Character Frequency Count

Start by counting the frequency of each character in string s. This can be done using a hash table, where the key is the character and the value is its count. The key observation is that t must be such that its characters divide evenly across s.

Check Divisibility of String Length

The length of string t must divide the length of s. Therefore, we check all divisors of the length of s and try to form an anagram of t using the available character frequencies. The correct length of t is the smallest divisor that allows this.

Form Anagram and Validate

For each divisor, attempt to construct string t using the available character counts. Ensure that the frequencies of the characters in s can be evenly divided by the divisor. The first valid t is the answer.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used for finding the divisors of the string length and counting character frequencies. Finding divisors can take O(sqrt(n)), where n is the length of s. Counting frequencies takes O(n). The overall complexity is O(n + sqrt(n)). Space complexity is O(1) if we only count frequencies, as the number of distinct characters is limited to 26.

What Interviewers Usually Probe

The candidate effectively uses hash tables for counting characters.
The candidate recognizes that the answer is a divisor of s.length.
The candidate efficiently checks possible divisors without unnecessary computation.

Common Pitfalls or Variants

Common pitfalls

Failing to recognize that the length of t must be a divisor of the length of s.
Overcomplicating the problem by checking all possible substrings of s instead of focusing on divisors.
Incorrectly counting character frequencies or not checking for even division of characters.

Follow-up variants

Instead of a string of lowercase letters, consider a string with uppercase or mixed case characters.
Introduce different constraints, like restricting the number of anagram groups.
Allow t to be longer or shorter and adjust the problem accordingly.

FAQ

How do I determine the length of string t?

The length of string t must be a divisor of the length of s, and t should allow all characters of s to be split evenly into anagrams of t.

What is the key observation for solving this problem?

The key observation is that the length of t must divide the length of s. This allows for efficient checking of possible candidates for t.

What is the time complexity of this problem?

The time complexity is O(n + sqrt(n)), where n is the length of s. The O(n) comes from counting character frequencies, and O(sqrt(n)) is for finding divisors of n.

Can the length of t be equal to the length of s?

Yes, the length of t can be equal to the length of s. In such a case, the string t could be the entire string s itself if s is already a valid anagram of t.

Why is hash table used here?

Hash tables are used to count the frequency of characters in string s. This allows for quick validation of whether a divisor length can form a valid string t.

terminal

Solution

Solution 1: Enumeration

Based on the problem description, the length of string $\textit{t}$ must be a factor of the length of string $\textit{s}$. We can enumerate the length $k$ of string $\textit{t}$ from small to large, and then check if it meets the requirements of the problem. If it does, we return. Thus, the problem is transformed into how to check whether the length $k$ of string $\textit{t}$ meets the requirements.

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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

Continue Practicing

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