How do you solve Count Anagrams on LeetCode?

Count Anagrams (LeetCode #2514) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding permutations requires knowledge of factorials.

What is the brute force solution for Count Anagrams?

The brute force approach for Count Anagrams is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of each word in the string and counting the distinct arrangements. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Anagrams?

Count Anagrams uses the following concepts: Hash Table, Math, String, Combinatorics, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Anagrams?

Explain your thought process clearly. Discuss time and space complexity upfront. Consider edge cases like single-letter words or repeated characters.

What are common mistakes when solving Count Anagrams?

Confusing permutations with combinations. Not accounting for duplicate characters in words.

#2514

Count Anagrams

Hard

Hash TableMathStringCombinatoricsCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of generating permutations, we can calculate the number of distinct anagrams mathematically using factorials and the frequency of characters. This avoids the inefficiency of generating permutations and allows us to compute the result in linear time.

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Algorithm

5 steps

1Step 1: Split the string into words.
2Step 2: For each word, count the frequency of each character.
3Step 3: Calculate the factorial of the length of the word.
4Step 4: Divide by the factorial of the frequency of each character to account for duplicates.
5Step 5: Multiply the results for all words and return the total modulo 10^9 + 7.

solution.py16 lines

1from collections import Counter
2from math import factorial
3
4MOD = 10**9 + 7
5
6def count_anagrams(s):
7    words = s.split()
8    total_anagrams = 1
9    for word in words:
10        freq = Counter(word)
11        word_length = len(word)
12        anagrams_count = factorial(word_length)
13        for count in freq.values():
14            anagrams_count //= factorial(count)
15        total_anagrams *= anagrams_count
16    return total_anagrams % MOD

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once to count characters and compute factorials. The space complexity is O(n) due to the storage of character frequencies.

1Understanding permutations requires knowledge of factorials.
2Character frequency is crucial for calculating distinct arrangements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.