Introduction

In this blog post, we will explore how to solve the LeetCode problem "242" 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

Given two strings s and t, return true if t is an anagram of s, and false otherwise. Example 1: Input: s = "anagram", t = "nagaram" Output: true Example 2: Input: s = "rat", t = "car" Output: false Constraints: 1 <= s.length, t.length <= 5 * 104 s and t consist of lowercase English letters. Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Explanation

Here's a breakdown of the solution:

• Character Frequency Counting: The core idea is to count the frequency of each character in both strings. If the frequencies are identical, the strings are anagrams.
• Hash Map (Dictionary): A hash map (dictionary in Python) is used to store the character counts efficiently.

• Comparison: Finally, the character counts are compared. If any character has different counts in the two strings, they're not anagrams.

• Runtime Complexity: O(n), where n is the length of the strings. Storage Complexity: O(1) because we're dealing with lowercase English letters, so the size of hash map is bounded by a small constant (26). For Unicode characters, the space complexity becomes O(k) where k is number of unique unicode characters.

Code

    def is_anagram(s: str, t: str) -> bool:
    """
    Given two strings s and t, return true if t is an anagram of s, and false otherwise.

    Example 1:
    Input: s = "anagram", t = "nagaram"
    Output: true

    Example 2:
    Input: s = "rat", t = "car"
    Output: false

    Constraints:
    1 <= s.length, t.length <= 5 * 104
    s and t consist of lowercase English letters.
    """

    if len(s) != len(t):
        return False

    s_char_counts = {}
    t_char_counts = {}

    for char in s:
        s_char_counts[char] = s_char_counts.get(char, 0) + 1

    for char in t:
        t_char_counts[char] = t_char_counts.get(char, 0) + 1

    return s_char_counts == t_char_counts