# Group Anagrams

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

An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

```
class Solution:
def groupAnagrams(self, strs):
anagram_map = defaultdict(list)
for word in strs:
sorted_word = ''.join(sorted(word))
anagram_map[sorted_word].append(word)
return list(anagram_map.values())
```

In this video instead of using the sorted word, he manually builds up a key based on the counts of the characters in the word. This is fine but there isn't any benefit to doing it this way. It's actually slower.

## Related Problems

Given a string **S** and an integer **K**, return *the length of the longest substring of* **S** *that contains at most* **K** **distinct*** characters*.

Given an integer array nums, you need to find one **continuous subarray** such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.

Return *the shortest such subarray and output its length*.

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals* after the insertion*.

**Note** that you don't need to modify intervals in-place. You can make a new array and return it.

Given two binary strings a and b, return *their sum as a binary string*. ie "01" and "10" as strings add to "11"