Group Anagrams LeetCode Problem 49 [Python Solution]

Welcome back! Today, we’re diving into a classic coding problem: the Group Anagrams LeetCode problem.

Anagrams are words or phrases formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

Our task is to take an array of strings and group the anagrams together.

Question

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

Problem Overview

We’re given an array of strings strs, and we need to group anagrams together. An efficient solution is required, as the input size can be substantial.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Understanding the Constraints

Before we worry about a solution, we need to understand the problem first.

So, let’s take a moment to understand the problem constraints:

• We can have anywhere from 1 to 10,000 strings in the input array.
• Each string can be as short as 0 characters or as long as 100 characters.
• All the characters in the strings are lowercase English letters.

Now that we’ve got the lay of the land, let’s explore a brute-force approach before diving into our efficient solution.

Group Anagrams LeetCode Problem Solution

1. Bruteforce Approach

A straightforward approach to solving this problem is sorting each string and grouping the sorted strings together.

However, this approach has a time complexity of O(m * n * log(n)), where m is the number of strings, and n is the average length of each string.

Sorting each string takes O(n * log(n)) time, and we must do this for m strings.

2. An Efficient Approach with a Hash Map

To achieve an efficient solution, we’ll use a hashmap.

Since the input strings consist of lowercase English letters, we can use an array of size 26 (one for each character 'a' to 'z') to represent the character count of each string.

Let’s break down the efficient approach step by step:

1. Initialize an empty hashmap called result. This hashmap will map a tuple of character counts to a list of anagrams.
2. For each string string in the input strs, create an array called count, initially filled with zeros.
3. Iterate through each character in the string string, and for each character character, increment the count at the corresponding index in the count array. To map characters ‘a’ to ‘z’ to indices 0 to 25, subtract the ASCII value of ‘a’ from the ASCII value of character.
4. After counting the characters in the string string, convert the count array into a tuple and use it as the key in the result hashmap. Append the string string to the list associated with that key.
5. If the key (tuple) doesn’t exist in the result hashmap, create it with an empty list as the value.
6. Finally, return the values (lists of anagrams) from the result hashmap.

Group Anagrams Python Code Solution

Here’s the Python code for the efficient solution:

def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
    answer = collections.defaultdict(list)

    for string in strs:
        count = [0] * 26
        for character in string:
            count[ord(character) - ord("a")] += 1
        answer[tuple(count)].append(string)
    return answer.values()

Time and Space Complexity

• O(m * n)where m is the number of strings and n is the average length of each string.
• O(m)

This efficient solution offers a stellar performance with its time complexity of O(m * n), where m represents the number of strings, and n stands for the average length of each string.

The reason for this efficiency is that we traverse each character of every string only once, and the operations performed for each character are constant-time.

As for space complexity, it’s worth noting that we utilize a hashmap (result) to store the grouping of anagrams. In the worst-case scenario, where there are no anagrams, this hashmap would have m keys, each mapping to a list of strings. Therefore, the space complexity is O(m), considering the additional space required for the count array, which has a constant size of 26.

In practice, this algorithm is highly efficient and performs exceptionally well even with large inputs. It’s a shining example of how clever data structures and careful manipulation of data can lead to optimal solutions in the world of coding.

Now that we’ve thoroughly dissected and conquered the challenge of grouping anagrams, you’re armed with another valuable tool in your coding arsenal. Keep exploring, keep innovating, and keep coding!

Reasoning Behind Our Approach

Our efficient approach leverages that anagrams have the same character counts when considering lowercase English letters.

By counting the characters in each string and using these counts as keys in a hashmap, we can efficiently group anagrams together without the need for sorting.

In summary, we’ve explored a problem that involves grouping anagrams and provided an efficient Python solution that utilizes a hashmap to achieve this grouping.

The solution is optimized for both time and space complexity, making it suitable for large inputs.

I hope you found this guide helpful, especially if you’re new to solving coding problems. If you have any questions or would like to see more coding guides, please let me know.

Solve Group Anagrams on LeetCode now.

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Happy coding!