Introduction

In this blog post, we will explore how to solve the LeetCode problem "49" 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 an array of strings strs, group the anagrams together. You can return the answer in any order. Example 1: Input: strs = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]] Explanation: There is no string in strs that can be rearranged to form "bat". The strings "nat" and "tan" are anagrams as they can be rearranged to form each other. The strings "ate", "eat", and "tea" are anagrams as they can be rearranged to form each other. Example 2: Input: strs = [""] Output: [[""]] Example 3: Input: strs = ["a"] Output: [["a"]] Constraints: 1 <= strs.length <= 104 0 <= strs[i].length <= 100 strs[i] consists of lowercase English letters.

Explanation

Here's the approach to solve the anagram grouping problem:

• Key Generation: Generate a unique key for each string that represents its anagram group. A sorted string of characters works well as the key since anagrams will have the same sorted string.
• Grouping: Use a dictionary (hash map) to store the anagrams. The key of the dictionary will be the generated key (sorted string), and the value will be a list of strings belonging to that anagram group.

• Result: Convert the dictionary values (lists of anagrams) into the final result list.

• Runtime Complexity: O(n * k log k), where n is the number of strings and k is the maximum length of a string. The log k factor comes from sorting each string.

• Storage Complexity: O(n * k) in the worst case, where n is the number of strings and k is the average length of the strings (for storing the hashmap and the result).

Code

    def groupAnagrams(strs):
    """
    Groups anagrams together from a list of strings.

    Args:
        strs: A list of strings.

    Returns:
        A list of lists, where each inner list contains anagrams.
    """

    anagram_groups = {}
    for s in strs:
        key = "".join(sorted(s))  # Generate key by sorting the string
        if key in anagram_groups:
            anagram_groups[key].append(s)
        else:
            anagram_groups[key] = [s]

    return list(anagram_groups.values())