DSA Walkthrough - 04

Intro

• The problem is, a bunch of strings will be give in an array, we need to we need to group the anagrams together.
• This is a both grouping and anagram problem

First approach

• My first approach was the classic hashmap approach, we store them together with their count and print them.
• What i thought at first was that we can check the len of the words present and if they are same then we can say that the given word is an anagram
• The major different between this and the regular anagram problem is that in regular anagram we just return the bool if the given word is an anagram.
• But in here, we cant do that as the list may contain the same len of letters but they can be any different alphabets.
• Therefore with the help of my close friend (ChatGPT), I was able to figure out that we cant really compare the len, instead we sort them individually and store them in the hashmap.

Solution

• The solution goes pretty much like this,.
• we create an hashmap, we loop through the array.
• We then sort every str from strs and join them to form the sorted word
• Now we use the sorted word as the key for the hash and the original str become the values of the hash
• we can use dictionarydict in python that automatically creates key and empty value when the key is absent
• Or we can create the key manually by using the if statements.
• in the end we return them as list of hash values

Code

class Solution:

    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
		hsh = {}        
        for s in strs:
            key = ''.join(sorted(s))
            if key not in hsh:
	            hsh[key] = []
            hsh[key].append(s)
            
        return list(hsh.values())

Conclusion

This was a fun problem to solve and as simple as it seems, it was a problem to think and solve.