What is the time complexity of Minimum Length of Anagram Concatenation?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Length of Anagram Concatenation?

This problem uses Hash Table, String, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Length of Anagram Concatenation?

Insight 1: The length of 't' must be a divisor of the length of 's'. This is because 's' is formed by concatenating anagrams of 't'. Insight 2: If a candidate length 'k' (length of t) is a divisor of the length of 's', then we need to check if 's' can be divided into chunks of length 'k' such that all chunks are anagrams of each other.

Which companies ask LeetCode 3138. Minimum Length of Anagram Concatenation in interviews?

This problem is frequently asked at Turing, UKG.

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Minimum Length of Anagram Concatenation - LeetCode 3138 Solution

Minimum Length of Anagram Concatenation - Complete Solution Guide

Minimum Length of Anagram Concatenation is LeetCode problem 3138, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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: Inp

Detailed Explanation

The problem asks us to find the minimum possible length of a string 't' such that the given string 's' is a concatenation of anagrams of 't'. An anagram is a rearrangement of letters. The input is the string 's', and the output is the minimum length of 't'. For instance, if s = 'abba', then t could be 'ba' (length 2), because 'abba' can be seen as 'ba' and 'ab', where 'ab' is an anagram of 'ba'.

Solution Approach

The provided solution iterates through all possible lengths 'k' of 't' from 1 to the length of 's'. For each 'k', it checks if 'k' is a divisor of the length of 's'. If it is, the algorithm calculates the character counts of the first 'k' characters of 's'. Then, it iterates through the remaining chunks of length 'k' in 's' and compares their character counts with the character counts of the first chunk. If all chunks have the same character counts (i.e., they are anagrams of each other), then 'k' is a valid candidate for the minimum length, and the algorithm returns 'k'. If no such 'k' is found, it means that 's' itself is the smallest possible string 't', so the algorithm returns the length of 's'.

Step-by-Step Algorithm

Step 1: Get the length 'n' of the string 's'.
Step 2: Iterate through all possible lengths 'k' from 1 to 'n'.
Step 3: For each 'k', check if 'n' is divisible by 'k'. If not, continue to the next 'k'.
Step 4: If 'n' is divisible by 'k', calculate the character counts of the first 'k' characters of 's' and store them in an array `counts1`.
Step 5: Iterate through the remaining chunks of length 'k' in 's'. For each chunk, calculate its character counts and store them in `counts2`.
Step 6: Compare `counts1` and `counts2`. If they are not equal, then the chunks are not anagrams of each other. Set `is_possible` to `false` and break out of the inner loop.
Step 7: If the inner loop completes without setting `is_possible` to `false`, then all chunks are anagrams of each other. Return 'k'.
Step 8: If the outer loop completes without finding a valid 'k', then return 'n'.

Key Insights

Insight 1: The length of 't' must be a divisor of the length of 's'. This is because 's' is formed by concatenating anagrams of 't'.
Insight 2: If a candidate length 'k' (length of t) is a divisor of the length of 's', then we need to check if 's' can be divided into chunks of length 'k' such that all chunks are anagrams of each other.
Insight 3: We can use character counts to efficiently check if two strings are anagrams. Two strings are anagrams if they have the same character counts.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Turing, UKG.

Minimum Length of Anagram Concatenation | LeetCode 3138

Medium

Hash Table String Counting

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minAnagramLength(String s) {
        int n = s.length();
        for (int k = 1; k <= n; k++) {
            if (n % k == 0) {
                int[] counts1 = new int[26];
                for (int i = 0; i < k; i++) {
                    counts1[s.charAt(i) - 'a']++;
                }

                boolean is_possible = true;
                for (int i = k; i < n; i += k) {
                    int[] counts2 = new int[26];
                    for (int j = 0; j < k; j++) {
                        counts2[s.charAt(i + j) - 'a']++;
                    }

                    if (!java.util.Arrays.equals(counts1, counts2)) {
                        is_possible = false;
                        break;
                    }
                }

                if (is_possible) {
                    return k;
                }
            }
        }
        return n;
    }
}

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Problem Info

Description

Problem 3138: Minimum Length of Anagram Concatenation - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

TuringUKG

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3138

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Get the length 'n' of the string 's'.
2
Step 2: Iterate through all possible lengths 'k' from 1 to 'n'.
3
Step 3: For each 'k', check if 'n' is divisible by 'k'. If not, continue to the next 'k'.
4
Step 4: If 'n' is divisible by 'k', calculate the character counts of the first 'k' characters of 's' and store them in an array `counts1`.
5
Step 5: Iterate through the remaining chunks of length 'k' in 's'. For each chunk, calculate its character counts and store them in `counts2`.
6
Step 6: Compare `counts1` and `counts2`. If they are not equal, then the chunks are not anagrams of each other. Set `is_possible` to `false` and break out of the inner loop.
7
Step 7: If the inner loop completes without setting `is_possible` to `false`, then all chunks are anagrams of each other. Return 'k'.
8
Step 8: If the outer loop completes without finding a valid 'k', then return 'n'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates up to 'n' times. The inner loop iterates 'n/k' times, where k can be close to 1, leading to O(n) complexity. Comparing two character count arrays of size 26 takes O(1). In total, the algorithm takes O(n * n) time which simplifies to O(n^2). The algorithm uses a fixed-size array of size 26 to store character counts, which takes O(1) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Rolling Hash

A rolling hash function could be used to efficiently calculate the hash of each substring of length 'k' and compare it with the hash of the first substring. This can potentially improve the average-case time complexity.

Time: O(n) on average with good hash function, O(n^2) worst caseSpace: O(1)

Trade-offs: Rolling hash can offer better performance on average, especially when the anagrams are shorter and more frequent. However, hash collisions can lead to O(n^2) complexity in the worst case and need to be handled carefully.

✨

Key Insights

1Insight 1: The length of 't' must be a divisor of the length of 's'. This is because 's' is formed by concatenating anagrams of 't'.
2Insight 2: If a candidate length 'k' (length of t) is a divisor of the length of 's', then we need to check if 's' can be divided into chunks of length 'k' such that all chunks are anagrams of each other.
3Insight 3: We can use character counts to efficiently check if two strings are anagrams. Two strings are anagrams if they have the same character counts.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: When the input string 's' consists of only one character repeated multiple times, the minimum length is 1. The algorithm handles this case correctly because it will find that k=1 is a divisor of n and all substrings of length 1 are anagrams of each other.
Edge case 2: When the input string 's' is already an anagram of itself (e.g., 'cdef'), the algorithm returns the length of 's', which is the correct answer. This happens when no smaller 'k' satisfies the anagram condition.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if 'k' is a divisor of the length of 's'. If 'k' is not a divisor, the string 's' cannot be formed by concatenating anagrams of length 'k'.

✗Common mistake 2: Incorrectly comparing the character counts. Ensure that the character counts are compared for all 26 characters and not just for some characters.

✗Common mistake 3: Assuming that the minimum length is always 1. If no 'k' less than n works, it is possible the minimum length is the length of s itself.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: In DNA sequencing, this algorithm could be adapted to find the shortest repeating sequence that makes up a larger DNA strand, where mutations are considered as anagrams.

🏦Real-world scenario 2: In data compression, this can be used to find repeating patterns or motifs in data streams. Identifying such repeating patterns allows to compress data representing larger sequences of data in terms of smaller repeated anagrams.

🎮Real-world scenario 3: Analyzing network traffic patterns to find repeating sequences of packets, which can indicate malicious activity or inefficient network usage. Finding a minimal repeating pattern of network packets could help understand communication structures.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Length of Anagram Concatenation"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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