49 GROUP ANAGRAMS.rtf

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\f0\fs24 \cf0 49. GROUP ANAGRAM \
\
\pard\pardeftab560\slleading20\partightenfactor0

\f1\fs26 \cf0 #GROUP ANAGRAM\
class Solution:\
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:\
        Hashmap = defaultdict(list)\
        for words in strs:\
            count = [0]*26\
            for i in range(len(words)):\
                count[ord(words[i])-ord('a')] += 1\
            \
            # two cases if the word already in hashmap or if the word not in hashmap     \
            \
            if tuple(count) not in Hashmap:\
                 #append the new list\
                Hashmap[tuple(count)] =  [words]\
            else:\
                #update add the words in the list\
                Hashmap[tuple(count)].append(words)\
\
            #Hashmap[tuple(count)].append(words)\
            \
        \
        return Hashmap.values()\
\pard\pardeftab560\slleading20\pardirnatural\partightenfactor0
\cf0 \
\pard\pardeftab560\slleading20\partightenfactor0
\cf0 # Neetcode approach \
\
        dicti = defaultdict(list) # countofcharacterarray : [list of words that matches count]\
\
        for s in strs:\
            count = [0] * 26\
\
            for c in s:\
\
                #generate keys \
                count[ord(c)-ord('a')] +=1\
\
            #adding word to hashmap dicti\
            #note: list cannot be keys so convert \
            #note: what if key not present that is why we use default dict \
            dicti[tuple(count)].append(s)\
        \
        return dicti.values()\
}