Minimum Length of Anagram Concatenation

Minimum Length of Anagram Concatenation

1. What is this problem about?

The "Minimum Length of Anagram Concatenation" problem asks you to find the smallest possible length of a string t such that the original string s is a concatenation of anagrams of t. An anagram of t is a string that contains the exact same characters as t but potentially in a different order.

For example, if s = "abba", the minimum length is 2 because s can be seen as "ab" followed by "ba", and "ba" is an anagram of "ab". If s = "abcabc", the minimum length could be 3 ("abc" + "abc") or even 6 ("abcabc"). We want the smallest such length.

2. Why is this asked in interviews?

This problem tests a candidate's ability to combine String manipulation with Number Theory (Divisors). It requires checking potential lengths efficiently. Interviewer's look for:

• Logical Deduction: Realizing that the length of the base anagram k must be a divisor of the total length n.
• Frequency Analysis: Understanding that for a length k to be valid, the total counts of each character in s must all be divisible by n/k.
• Complexity Management: Avoiding unnecessary checks by iterating only through divisors.

3. Algorithmic pattern used

The pattern is Divisor Iteration + Frequency Counting.

1. Find all divisors of the length of the string s.
2. For each divisor k (from smallest to largest):
   • Check if k could be a valid anagram length.
   • For a length k to be valid, if we split the string into blocks of size k, every block must have the same character frequency.
   • A faster check: The total count of each character in the entire string s must be a multiple of the number of blocks (n/k).

4. Example explanation

String: s = "cabbac" (length 6) Divisors: 1, 2, 3, 6 Total counts: a: 2, b: 2, c: 2

• Try length 1: Blocks are "c", "a", "b", "b", "a", "c". Not anagrams.
• Try length 2: Blocks are "ca", "bb", "ac". "ca" and "bb" are not anagrams.
• Try length 3: Blocks are "cab", "bac". "bac" is an anagram of "cab". Min length is 3.

5. Common mistakes candidates make

• Checking all lengths: Iterating from 1 to n instead of only divisors.
• Incorrect Anagram Check: Using sorting to check anagrams inside the loop, which adds $O(n \log n)$ per check. Using a frequency array of size 26 is much faster $O(26)$.
• Ignoring character counts: Not realizing that character frequencies in the first block must match character frequencies in every other block.

6. Interview preparation tip

When you see a problem where a string is composed of equal-sized "blocks" or "parts," always look at the divisors of the total length. This is a common trick to reduce the search space.