Introduction

In this blog post, we will explore how to solve the LeetCode problem "2514" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 ith word of t is a permutation of the ith 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 109 + 7. Example 1: Input: s = "too hot" Output: 18 Explanation: Some of the anagrams of the given string are "too hot", "oot hot", "oto toh", "too toh", and "too oht". Example 2: Input: s = "aa" Output: 1 Explanation: There is only one anagram possible for the given string. Constraints: 1 <= s.length <= 105 s consists of lowercase English letters and spaces ' '. There is single space between consecutive words.

Explanation

Here's the breakdown of the solution:

• Word-by-Word Anagram Calculation: Split the input string into individual words. For each word, calculate the number of distinct anagrams it can form.
• Factorial and Modular Arithmetic: The number of anagrams for a word is calculated using factorials. To handle large numbers, perform all calculations modulo 10⁹ + 7. Use the formula: n! / (freq1! * freq2! * ...) where n is the word length, and freq1, freq2... are the frequencies of each character in the word.

• Modular Inverse Optimization: Efficiently compute the modular inverse required for division in the modulo space using Fermat's Little Theorem. This avoids floating-point operations and maintains precision.

• Runtime Complexity: O(N), where N is the length of the input string s.

• Storage Complexity: O(M), where M is the maximum length of a word in the string s.

Code

    def num_distinct_anagrams(s: str) -> int:
    """
    Calculates the number of distinct anagrams of a string where each word's anagram
    must stay in its original position.

    Args:
        s: The input string containing words separated by single spaces.

    Returns:
        The number of distinct anagrams modulo 10^9 + 7.
    """

    MOD = 10**9 + 7

    def factorial(n):
        """Calculates n! modulo MOD."""
        result = 1
        for i in range(1, n + 1):
            result = (result * i) % MOD
        return result

    def power(base, exp, mod):
        """Calculates (base^exp) % mod using binary exponentiation."""
        result = 1
        base %= mod
        while exp > 0:
            if exp % 2 == 1:
                result = (result * base) % mod
            base = (base * base) % mod
            exp //= 2
        return result

    def mod_inverse(n, mod):
        """Calculates the modular inverse of n modulo mod using Fermat's Little Theorem."""
        return power(n, mod - 2, mod)

    words = s.split()
    total_anagrams = 1

    for word in words:
        n = len(word)
        counts = {}
        for char in word:
            counts[char] = counts.get(char, 0) + 1

        numerator = factorial(n)
        denominator = 1
        for count in counts.values():
            denominator = (denominator * factorial(count)) % MOD

        anagrams = (numerator * mod_inverse(denominator, MOD)) % MOD
        total_anagrams = (total_anagrams * anagrams) % MOD

    return total_anagrams