## 49. Group Anagrams

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:

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

Note: All inputs will be in lower-case.

b'
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#### Approach #1: Categorize by Sorted String [Accepted]

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Intuition

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Two strings are anagrams if and only if their sorted strings are equal.

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Algorithm

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Maintain a map ans : {String -> List} where each key is a sorted string, and each value is the list of strings from the initial input that when sorted, are equal to .

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In Java, we will store the key as a string, eg. code. In Python, we will store the key as a hashable tuple, eg. (\'c\', \'o\', \'d\', \'e\').

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Complexity Analysis

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Time Complexity: , where is the length of strs, and is the maximum length of a string in strs. The outer loop has complexity as we iterate through each string. Then, we sort each string in time.

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Space Complexity: , the total information content stored in ans.

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#### Approach #2: Categorize by Count [Accepted]

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Intuition

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Two strings are anagrams if and only if their character counts (respective number of occurrences of each character) are the same.

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Algorithm

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We can transform each string into a character count, , consisting of 26 non-negative integers representing the number of \'s, \'s, \'s, etc. We use these counts as the basis for our hash map.

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In Java, the hashable representation of our count will be a string delimited with \'#\' characters. For example, abbccc will be #1#2#3#0#0#0...#0 where there are 26 entries total. In python, the representation will be a tuple of the counts. For example, abbccc will be (1, 2, 3, 0, 0, ..., 0), where again there are 26 entries total.

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Complexity Analysis

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Time Complexity: , where is the length of strs, and is the maximum length of a string in strs. Counting each string is linear in the size of the string, and we count every string.

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Space Complexity: , the total information content stored in ans.

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Analysis written by: @awice

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