## Problem Statement

Given a list of strings, the task is to group the anagrams together. 

An anagram is a word or phrase formed by rearranging the letters of another, such as "cinema", formed from "iceman".

## Example Generation

### Example 1:
- **Input:** `["dog", "god", "hello"]`
- **Output:** `[["dog", "god"], ["hello"]]`
- **Justification:** "dog" and "god" are anagrams, so they are grouped together. "hello" does not have any anagrams in the list, so it is in its own group.

### Example 2:
- **Input:** `["listen", "silent", "enlist"]`
- **Output:** `[["listen", "silent", "enlist"]]`
- **Justification:** All three words are anagrams of each other, so they are grouped together.

### Example 3:
- **Input:** `["abc", "cab", "bca", "xyz", "zxy"]`
- **Output:** `[["abc", "cab", "bca"], ["xyz", "zxy"]]`
- **Justification:** "abc", "cab", and "bca" are anagrams, as are "xyz" and "zxy".

**Constraints:**

- 1 <= strs.length <= 10<sup>4</sup>
- `0 <= strs[i].length <= 100`
- `strs[i]` consists of lowercase English letters.

## Try it yourself
Try solving this question here:


## Problem Statement

Given a list of strings, the task is to group the anagrams together. 

An anagram is a word or phrase formed by rearranging the letters of another, such as "cinema", formed from "iceman".

## Example Generation

### Example 1:
- **Input:** `["dog", "god", "hello"]`
- **Output:** `[["dog", "god"], ["hello"]]`
- **Justification:** "dog" and "god" are anagrams, so they are grouped together. "hello" does not have any anagrams in the list, so it is in its own group.

### Example 2:
- **Input:** `["listen", "silent", "enlist"]`
- **Output:** `[["listen", "silent", "enlist"]]`
- **Justification:** All three words are anagrams of each other, so they are grouped together.

### Example 3:
- **Input:** `["abc", "cab", "bca", "xyz", "zxy"]`
- **Output:** `[["abc", "cab", "bca"], ["xyz", "zxy"]]`
- **Justification:** "abc", "cab", and "bca" are anagrams, as are "xyz" and "zxy".

## Solution

- **Sorting Approach:**
  - For each word in the input list:
    - Sort the letters of the word.
    - Use the sorted word as a key in a hash map, and add the original word to the list of values for that key.
  - The hash map values will be the groups of anagrams.

- **Why This Will Work:**
  - Anagrams will always result in the same sorted word, so they will be grouped together in the hash map.


### Algorithm Walkthrough

- Given the input `["abc", "cab", "bca", "xyz", "zxy"]`
- For "abc":
  - Sorted word is "abc".
  - Add "abc" to the hash map with key "abc".
- For "cab":
  - Sorted word is "abc".
  - Add "cab" to the list in the hash map with key "abc".
- Continue this process for all words.
- The hash map values are the groups of anagrams.

## Code


## Complexity Analysis

- **Time Complexity:** O(n*k*log(k)), where n is the number of strings, and k is the maximum length of a string in strs. This is because each of the n strings is sorted in O(k*log(k)) time.
- **Space Complexity:** O(n*k), where n is the number of strings, and k is the maximum length of a string in strs. This space is used for the output data structure and the hash map.


## Problem Statement

Given a list of strings, the task is to group the anagrams together. 

An anagram is a word or phrase formed by rearranging the letters of another, such as "cinema", formed from "iceman".

## Example Generation

### Example 1:
- **Input:** `["dog", "god", "hello"]`
- **Output:** `[["dog", "god"], ["hello"]]`
- **Justification:** "dog" and "god" are anagrams, so they are grouped together. "hello" does not have any anagrams in the list, so it is in its own group.

### Example 2:
- **Input:** `["listen", "silent", "enlist"]`
- **Output:** `[["listen", "silent", "enlist"]]`
- **Justification:** All three words are anagrams of each other, so they are grouped together.

### Example 3:
- **Input:** `["abc", "cab", "bca", "xyz", "zxy"]`
- **Output:** `[["abc", "cab", "bca"], ["xyz", "zxy"]]`
- **Justification:** "abc", "cab", and "bca" are anagrams, as are "xyz" and "zxy".

## Try it yourself
Try solving this question here:
