# Group Anagrams (Leetcode  #49)

### Question:

[Given an array of strings `strs`, group](https://leetcode.com/problems/group-anagrams/) the anagrams together. You can return the answer in any order.

An Anagram is a wor[d or phrase formed](https://leetcode.com/problems/group-anagrams/) by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

**Constraints:**

* [`1 <= strs.length <= 10<sup>4</sup>`](https://leetcode.com/problems/group-anagrams/)
    
* [`0 <= strs[`](https://leetcode.com/problems/group-anagrams/)`i].length` [`<= 100`](https://leetcode.com/problems/group-anagrams/)
    
* [`strs[i]`](https://leetcode.com/problems/group-anagrams/) consists of [lowercase English](https://leetcode.com/problems/group-anagrams/) letters.
    

### Answer:

1. Create a hashmap to keep track of each word where the key is the sorted word and the values will be the original word
    
2. Return the values of the hashmap in a list form
    

```python
word = {}
for w in strs:
    sorted_string = "".join(sorted(w))
    if sorted_string not in word:
        word[sorted_string] = []
    word[sorted_string].append(w)
return list(word.values())
```

**Time complexity**

Let's break this down into each part.

1. Loop each item in strs will cost you O(N)
    
2. Sorting string with average length K will cost you O(klogk)
    
3. Hence the overall time complexity will be O(n \* k log k)
    

**Space Complexity**

1. When we sort the text we need a temporary space of O(k)
    
2. We need to store each sorted string in the hashmap
    
3. The overall space complexity will be O(n \* k)