What is the time complexity of Count Anagrams?

The optimal solution has a time complexity of O(N + M).

What data structures are used to solve Count Anagrams?

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

What are the key insights for solving Count Anagrams?

Insight 1: Each word in the string `s` can be permuted independently to form different anagrams of `s`. Therefore, we can find the number of anagrams for each word and multiply them together. Insight 2: To count the number of anagrams for a word, we need to account for repeated characters. If a word has repeated characters, we must divide the total number of permutations (n!) by the factorial of the count of each repeated character.

Which companies ask LeetCode 2514. Count Anagrams in interviews?

This problem is frequently asked at MathWorks.

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Count Anagrams - LeetCode 2514 Solution

Count Anagrams - Complete Solution Guide

Count Anagrams is LeetCode problem 2514, a Hard 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 containing one or more words. Every consecutive pair of words is separated by a single space ' ' . A string t is an anagram of string s if the i th word of t is a permutation of the i th word of s . For example, "acb dfe" is an anagram of "abc def" , but "def cab" and "adc bef" are not. Return the number of distinct anagrams of s . Since the answer may be very large, return it modulo 10 9 + 7 . Example 1: Input: s = "too hot" Output: 18 Explanation: Some of the anagrams

Detailed Explanation

The problem asks us to count the number of distinct anagrams of a given string `s`, where `s` consists of words separated by single spaces. An anagram of `s` is formed by rearranging the letters within each word while preserving the order of the words. Since the number of anagrams can be very large, we need to return the result modulo 10^9 + 7. The constraints state that the string `s` has a length between 1 and 10^5, contains only lowercase English letters and spaces, and has single spaces between words.

Solution Approach

The solution involves the following steps: 1. Split the string `s` into individual words. 2. For each word, calculate the number of its anagrams. This involves: a. Calculating the factorial of the word's length. b. Counting the occurrences of each character in the word. c. Dividing the factorial by the factorial of the count of each character (using modular inverse to avoid actual division). 3. Multiply the number of anagrams of each word together (modulo 10^9 + 7). 4. Return the final result.

Step-by-Step Algorithm

Step 1: Split the input string `s` into an array of words using the space character as a delimiter.
Step 2: Calculate the maximum length of a word. This will be used to precompute the factorials up to that maximum length for efficiency.
Step 3: Precompute factorials modulo MOD (10^9 + 7) from 0 up to the maximum word length. Store these in an array `fact`.
Step 4: Define a helper function `modInverse(n)` that calculates the modular multiplicative inverse of `n` modulo MOD using Fermat's Little Theorem (n^(MOD-2) mod MOD).
Step 5: Initialize the answer `ans` to 1.
Step 6: Iterate through each word in the array of words.
Step 7: For each word: a. Calculate the length `n` of the word. b. Initialize `word_perms` to `fact[n]` (the factorial of the word length). c. Count the occurrences of each character in the word using a hash map. d. Iterate through the counts of each character in the hash map. e. For each character count, calculate the modular inverse of the factorial of the count using `modInverse(fact[count])`. f. Multiply `word_perms` by the modular inverse and take the modulo with MOD.
Step 8: Multiply `ans` by `word_perms` and take the modulo with MOD.
Step 9: Return `ans`.

Key Insights

Insight 1: Each word in the string `s` can be permuted independently to form different anagrams of `s`. Therefore, we can find the number of anagrams for each word and multiply them together.
Insight 2: To count the number of anagrams for a word, we need to account for repeated characters. If a word has repeated characters, we must divide the total number of permutations (n!) by the factorial of the count of each repeated character.
Insight 3: Since we need to perform division modulo 10^9 + 7, we need to calculate the modular inverse of the factorials of the counts of the repeated characters. Fermat's Little Theorem is used for this purpose: a^(p-2) mod p is the modular inverse of a mod p, where p is a prime number (10^9 + 7 is prime).

Complexity Analysis

Time Complexity: O(N + M)

Space Complexity: O(M)

Topics

This problem involves: Hash Table, Math, String, Combinatorics, Counting.

Companies

Asked at: MathWorks.

Count Anagrams | LeetCode 2514

Hard

Hash Table Math String Combinatorics Counting

Time: O(N + M)Space: O(M)

Solution Code

Available in:

class Solution {
    public int countAnagrams(String s) {
        int MOD = (int)1e9 + 7;

        String[] words = s.split(" ");

        int max_word_len = 0;
        for (String word : words) {
            max_word_len = Math.max(max_word_len, word.length());
        }

        long[] fact = new long[max_word_len + 1];
        fact[0] = 1;
        for (int i = 1; i <= max_word_len; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        
        
        long ans = 1;
        for (String word : words) {
            int n = word.length();
            
            long word_perms = fact[n];
            
            java.util.Map<Character, Integer> counts = new java.util.HashMap<>();
            for (char char_val : word.toCharArray()) {
                counts.put(char_val, counts.getOrDefault(char_val, 0) + 1);
            }
            
            for (int count : counts.values()) {
                long inv_fact_count = modInverse(fact[count], MOD);
                word_perms = (word_perms * inv_fact_count) % MOD;
            }
            
            ans = (ans * word_perms) % MOD;
            
        }
        
        return (int)ans;
    }
    
    private long modInverse(long n, int MOD) {
        return power(n, MOD - 2, MOD);
    }
    
    private long power(long base, long exp, int MOD) {
        long res = 1;
        base = base % MOD;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % MOD;
            exp = exp >> 1;
            base = (base * base) % MOD;
        }
        return res;
    }
}

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Problem Info

Description

Problem 2514: Count Anagrams - Detailed explanation and solution approaches.

Topics

Hash Table Math String Combinatorics Counting

Companies

MathWorks

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2514

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

9 steps

1
Step 1: Split the input string `s` into an array of words using the space character as a delimiter.
2
Step 2: Calculate the maximum length of a word. This will be used to precompute the factorials up to that maximum length for efficiency.
3
Step 3: Precompute factorials modulo MOD (10^9 + 7) from 0 up to the maximum word length. Store these in an array `fact`.
4
Step 4: Define a helper function `modInverse(n)` that calculates the modular multiplicative inverse of `n` modulo MOD using Fermat's Little Theorem (n^(MOD-2) mod MOD).
5
Step 5: Initialize the answer `ans` to 1.
6
Step 6: Iterate through each word in the array of words.
7
Step 7: For each word: a. Calculate the length `n` of the word. b. Initialize `word_perms` to `fact[n]` (the factorial of the word length). c. Count the occurrences of each character in the word using a hash map. d. Iterate through the counts of each character in the hash map. e. For each character count, calculate the modular inverse of the factorial of the count using `modInverse(fact[count])`. f. Multiply `word_perms` by the modular inverse and take the modulo with MOD.
8
Step 8: Multiply `ans` by `word_perms` and take the modulo with MOD.
9
Step 9: Return `ans`.

💡

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 + M)

💾

Space Complexity

O(M)

📝

Why?

The time complexity is determined by the string splitting and iterating through each character in each word to compute its anagram permutations, combined with pre-computing factorials. The space complexity is primarily determined by the size of the factorial array and the character count map.

📖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

Precompute all modular inverses

Instead of calculating the modular inverse for each count of a character in a word, we can precompute all the modular inverses from 1 to the maximum word length. This could slightly improve the time complexity by avoiding repeated calculations of modular inverses.

Time: O(N + M), but with potentially lower constant factors.Space: O(M), for storing the precomputed modular inverses.

Trade-offs: This approach trades off a bit of extra space for potentially faster computation, especially if the modular inverse calculation is relatively expensive.

✨

Key Insights

1Insight 1: Each word in the string `s` can be permuted independently to form different anagrams of `s`. Therefore, we can find the number of anagrams for each word and multiply them together.
2Insight 2: To count the number of anagrams for a word, we need to account for repeated characters. If a word has repeated characters, we must divide the total number of permutations (n!) by the factorial of the count of each repeated character.
3Insight 3: Since we need to perform division modulo 10^9 + 7, we need to calculate the modular inverse of the factorials of the counts of the repeated characters. Fermat's Little Theorem is used for this purpose: a^(p-2) mod p is the modular inverse of a mod p, where p is a prime number (10^9 + 7 is prime).

📚

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: Empty string or string with only spaces. The provided code will work as expected since `s.split()` on an empty string returns an empty list and the loop will not execute. It will return 1 because `ans` is initialized to 1.
Edge case 2: String with only one word. The code correctly calculates the number of anagrams for that single word and returns the result.
Edge case 3: Word with repeating characters like 'aaaaa'. The algorithm correctly handles this using modular arithmetic and inverses.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. The problem requires returning the answer modulo 10^9 + 7. If you don't take the modulo at each multiplication and division step, the numbers can become too large and cause integer overflow, resulting in incorrect answers.

✗Common mistake 2: Using regular division instead of modular inverse when calculating the number of permutations with repeated characters. Direct division doesn't work in modular arithmetic; we need to use the modular inverse.

✗Common mistake 3: Not handling the edge case of empty string or string with only spaces correctly.

✅

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: Text analysis and natural language processing. Counting anagrams can be useful in analyzing the composition of text, identifying patterns, or generating variations of text for different purposes.

🏦Real-world scenario 2: Cryptography. While not directly related, the concepts of permutations and combinations, which are used in this problem, are fundamental to understanding cryptographic algorithms and techniques.

🎮Real-world scenario 3: Bioinformatics. Analyzing DNA sequences for specific patterns and variations, where permutations and combinations might be relevant.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: November 22, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Count Anagrams"?

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

🤖 Need help with this problem?