3138. Minimum Length of Anagram Concatenation

MediumHash TableStringCounting

Problem Description

You are given a string s, which is known to be a concatenation of anagrams of some string t. Your task is to 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".

Intuition

To solve this problem, note that the length of string t must be a factor of the length of string s, because the entire string s is constructed by repeating anagrams of string t. We aim to find the smallest possible k which is a divisor of the length n of s, such that s can be divided into chunks of size k where each chunk is an anagram of t.

Count Characters: First, we count the occurrences of each character in the string s and store these counts.
Check Factors: Traverse potential lengths k which are factors of n, starting from the smallest.
Verify Chunks: For each k, check if the entire string s can be subdivided into substrings of length k, each having the character distribution necessary to form an anagram of t. This means each chunk of length k should have the same character frequency such that when multiplied by the number of such chunks (n/k), it matches the frequency in s.

Whenever k satisfies the condition, it implies the smallest possible k and hence the minimum possible length of t.

Quick Interview Experience

Help others by sharing your interview experience

Have you seen this problem before?

Solution Approach

The problem is approached using enumeration to find the smallest possible length of t, which is a factor of the length of s. Here's how the solution works:

Character Counting: First, a character frequency map cnt for the entire string s is created using the Counter from the collections module. This map keeps track of how many times each character appears in s.
Iterate Over Possible Lengths: Iterate over each possible length k from 1 to n (n being the length of s). Only consider k if it is a divisor of n. This ensures we are checking only those lengths which can evenly divide s into parts.
Verification Function: Define a helper function check(k) to determine if k can be a valid length for t. In this function:
- Sub-Chunks Analysis: Iterate through the string s in segments of length k. For each segment, calculate the frequency of characters in that segment using a local Counter.
- Frequency Comparison: Compare each character's frequency across the full string s (in cnt) against the frequency expected from the current segment (cnt1). Specifically, verify that each character frequency in cnt1 multiplied by n/k (the number of such segments) matches the corresponding frequency in cnt.
- If all comparisons hold true for the current k, check(k) returns True.
Return the Minimum Length: As soon as a valid k is found such that check(k) returns True, return k as it represents the smallest possible length of t.

This approach efficiently finds the minimum possible length of t by leveraging divisor properties and frequency validation over potential substring partitions of s.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's walk through a small example using the solution approach described.

Assume the input string s is "abcabc". Our goal is to find the minimum possible length of a string t such that s is a concatenation of anagrams of t.

Character Counting: First, determine the character frequency for s. Here, s = "abcabc", so the frequency map is:
- 'a': 2
- 'b': 2
- 'c': 2
Iterate Over Possible Lengths: Next, iterate over each potential length for t. Given that s has length 6, the divisors are 1, 2, 3, and 6.
Check Each Length:
- k = 1: Can't form valid anagrams as single-character chunks can't rearrange to yield the original.
- k = 2: Divide s into chunks "ab", "ca", "bc". Check if each can be a rearrangement of a 2-character t. The character frequencies won't match across the full string, failing this check.
- k = 3: Divide s into chunks "abc", "abc". Each segment matches directly, perfectly forming the original string by rearrangement. Check if the character frequencies match: for each chunk, the local Counter would be {'a': 1, 'b': 1, 'c': 1}, and scaling this by n/k (which is 2) matches the full count in s.
Return the Minimum Length: For k = 3, the check is successful, indicating t could be "abc". This implies the minimum length of t is 3. Thus, it satisfies the condition, and 3 is returned as the result.

This illustrates how the solution efficiently finds the minimum possible length of t by examining divisor factors and their corresponding substring segmentations in s.

Solution Implementation

1from collections import Counter
2
3class Solution:
4    def minAnagramLength(self, s: str) -> int:
5        # Helper function to determine if the string can be divided into 
6        # anagrams of length `k`
7        def check(k: int) -> bool:
8            # Iterate over the string in chunks of size `k`
9            for i in range(0, n, k):
10                # Count the frequency of characters in each chunk
11                current_chunk_count = Counter(s[i : i + k])
12                # Check if count of each character multiplied by number of chunks equals the total count
13                for char, total_count in char_count.items():
14                    if current_chunk_count[char] * (n // k) != total_count:
15                        return False
16            return True
17
18        char_count = Counter(s)  # Count of all characters in the string
19        n = len(s)  # Length of the input string
20
21        # Check for each possible length `i` from 1 to `n` if the string can
22        # be divided evenly into segments of `i` that are anagrams
23        for i in range(1, n + 1):
24            # Proceed only if `n` is divisible by `i` and the chunks are valid anagrams
25            if n % i == 0 and check(i):
26                return i
27

1class Solution {
2    private int n; // Length of the input string
3    private char[] s; // Array of characters from the input string
4    private int[] cnt = new int[26]; // Frequency array for each letter in the alphabet
5
6    public int minAnagramLength(String s) {
7        n = s.length();
8        this.s = s.toCharArray();
9      
10        // Populate the cnt array with the frequency of each character in the input string
11        for (int i = 0; i < n; ++i) {
12            ++cnt[this.s[i] - 'a'];
13        }
14
15        // Iterate over possible lengths of subsequences
16        for (int i = 1; ; ++i) {
17            // Check if the current length is a divisor of n and if it can form an anagram
18            if (n % i == 0 && check(i)) {
19                return i; // Return the smallest divisor length that can form an anagram
20            }
21        }
22    }
23
24    private boolean check(int k) {
25        // Validate if the input string can be divided into blocks of size 'k' that are anagrams
26
27        for (int i = 0; i < n; i += k) {
28            int[] cnt1 = new int[26]; // Temporary frequency array for current block
29          
30            // Count the frequency of each character in the current block of size 'k'
31            for (int j = i; j < i + k; ++j) {
32                ++cnt1[s[j] - 'a'];
33            }
34          
35            // Compare the frequency of characters in the current block with the global frequency
36            for (int j = 0; j < 26; ++j) {
37                if (cnt1[j] * (n / k) != cnt[j]) {
38                    return false; // Return false if the frequencies do not match
39                }
40            }
41        }
42        return true; // Return true if all blocks are valid
43    }
44}
45

1class Solution {
2public:
3    int minAnagramLength(std::string s) {
4        int n = s.size();
5
6        // Count frequency of each character in the string
7        int charCount[26] = {};
8        for (char c : s) {
9            charCount[c - 'a']++;
10        }
11
12        // Lambda function to check if a division of length 'k' can form an anagram
13        auto check = [&](int k) {
14            // Traverse each segment of length 'k'
15            for (int i = 0; i < n; i += k) {
16                int segmentCount[26] = {};
17                for (int j = i; j < i + k; ++j) {
18                    segmentCount[s[j] - 'a']++;
19                }
20                // Verify if the segment's distribution matches with total string
21                for (int j = 0; j < 26; ++j) {
22                    if (segmentCount[j] * (n / k) != charCount[j]) {
23                        return false;
24                    }
25                }
26            }
27            return true;
28        };
29
30        // Iterate over possible lengths to find the minimum valid length
31        for (int i = 1;; ++i) {
32            if (n % i == 0 && check(i)) {
33                return i;
34            }
35        }
36    }
37};
38

1function minAnagramLength(s: string): number {
2    const n = s.length;
3    const count: Record<string, number> = {}; // Store the frequency of each character in the string
4
5    // Build a frequency count of the characters in the string
6    for (let i = 0; i < n; i++) {
7        count[s[i]] = (count[s[i]] || 0) + 1;
8    }
9
10    // Helper function to check if a given length k can divide the string into anagrams
11    const check = (k: number): boolean => {
12        for (let i = 0; i < n; i += k) {
13            const currentCount: Record<string, number> = {}; // Frequency of characters in the current segment
14
15            // Compute frequency for the current segment of length k
16            for (let j = i; j < i + k; j++) {
17                currentCount[s[j]] = (currentCount[s[j]] || 0) + 1;
18            }
19
20            // Verify if current segment can be part of the anagram structure
21            for (const [char, value] of Object.entries(count)) {
22                if (currentCount[char] * ((n / k) | 0) !== value) {
23                    return false; // If there's a mismatch, it's not valid
24                }
25            }
26        }
27        return true; // All segments are valid
28    };
29
30    // Start checking from the smallest possible segment length that divides n
31    for (let i = 1; ; ++i) {
32        if (n % i === 0 && check(i)) { // If n is divisible by i and it forms valid anagrams
33            return i; // Return the minimum length found
34        }
35    }
36}
37

Time and Space Complexity

The time complexity of the provided code is O(n \times A), where n is the length of the string s, and A is the number of divisors (factors) of n. This is because for each divisor i of n, the code checks if partitioning the string into n / i parts of length i can create a valid anagram structure. The space complexity is O(|\Sigma|), where \Sigma represents the character set, typically the set of lowercase letters, since it uses a Counter to tally the occurrences of characters in the string.

Learn more about how to find time and space complexity quickly.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

What are the two properties the problem needs to have for dynamic programming to be applicable? (Select 2)

Optimal substructure

Overlapping subproblems

Non-overlapping subproblems

Constant time subproblems