Group Anagrams (LC 49)

Hash Map Solution

The idea of this is basically to encode strings in map format by counting how many of each letter they have. For example, tar and rat have the same number of t’s, a’s, and r’s, so they would be encoded the same way. Then you can create a dictionary where the encoding is the key, and the value is how many strings with this encoding there are.

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        
        result = defaultdict(list)
 
        for s in strs:
            count = [0] * 26
 
            for c in s:
                count[ord(c) - ord("a")] += 1
 
            result[tuple(count)].append(s)
 
        return result.values()

Notes:

• defaultdict functions the same as dictionaries, but it never raises a KeyError and instead provides a default value. Here, we set the default value to a list.
• ord() gives the ASCII value of something, so ord("a")=65,ord("b)=66, etc.
  • This basically maps the alphabet from 1-26, just shifted up, but it doesn’t matter
• We use result[tuple(count)] instead of just result[count] because dictionary keys need to be immutable, so you can’t use an array