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# 49. Group Anagrams 🤹‍♂️

**Difficulty**: `Medium` - **Tags**: `Hash Table`, `String`, `Sorting`

[LeetCode Problem Link](https://leetcode.com/problems/group-anagrams/)

***

## Problem Statement 📜

Given an array of strings `strs`, group the anagrams together. An anagram is a word or phrase formed by rearranging the letters of another, using all the original letters exactly once.

The answer can be returned in any order.

***

## Examples 🌟

🔹 **Example 1:**

**Input:**

```plaintext
strs = ["eat", "tea", "tan", "ate", "nat", "bat"]
```

**Output:**

```plaintext
[["bat"], ["nat", "tan"], ["ate", "eat", "tea"]]
```

🔹 **Example 2:**

**Input:**

```plaintext
strs = [""]
```

**Output:**

```plaintext
[[""]]
```

🔹 **Example 3:**

**Input:**

```plaintext
strs = ["a"]
```

**Output:**

```plaintext
[["a"]]
```

***

## Constraints ⚙️

* `1 <= strs.length <= 10^4`
* `0 <= strs[i].length <= 100`
* `strs[i]` consists of lowercase English letters.

***

## Solution 💡

To group anagrams, we can use a `HashMap` where:

* The key is a representation of the sorted characters of a word.
* The value is a list of words that share the same sorted characters.

***

### Java Solution

```java
import java.util.*;

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> anagramMap = new HashMap<>();

        for (String str : strs) {
            char[] charArray = str.toCharArray();
            Arrays.sort(charArray);
            String sortedStr = new String(charArray);

            anagramMap.putIfAbsent(sortedStr, new ArrayList<>());
            anagramMap.get(sortedStr).add(str);
        }

        return new ArrayList<>(anagramMap.values());
    }
}
```

***

## Explanation of the Solution

1. **Sorting and Grouping**:
   * Convert each string to a character array, sort it, and convert it back to a string.
   * Use this sorted string as the key in a `HashMap`.
   * Add the original string to the list corresponding to this key.
2. **Result**:
   * The `HashMap` values contain grouped anagrams.
   * Return all the values as a list of lists.

***

## Time Complexity ⏳

* **O(n \* k log k)**:
  * Sorting each string (`k log k`, where `k` is the max length of a string).
  * Iterating through `n` strings.

## Space Complexity 💾

* **O(n \* k)**:
  * Space for the `HashMap` to store `n` strings of length `k`.

You can find the full solution [here](https://github.com/ChunhThanhDe/Leetcode-Top-Interview/blob/main/Topic%205%20Hashmap/043%20Group%20Anagrams/Solution.java).


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