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  1. Prepare
  2. Python
  3. Collections
  4. Word Order
  5. Discussions

Word Order

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1666 Discussions

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  • aamagantal
     + 0 comments
    from collections import OrderedDict
myDict = OrderedDict()
n = int(input())
for _ in range(n):
    word = str(input())
    myDict[word] = myDict.get(word, 0) + 1
print(len(myDict))
print(*myDict.values())

  • 771aravind
     + 0 comments

    Simple method using Counter

    from collections import Counter

    n = int(input()) strs=Counter() for _ in range(n): arr=input() strs[arr]+=1 print(len(strs)) print(*strs.values())

  • devesh_sharma
     + 0 comments

    Simple solution without counter, just dict

    if __name__ == '__main__':
    n = int(input())
    words_dict = {}
    for _ in range(n):
        word = input()
        words_dict[word] = 1 if word not in words_dict else words_dict[word] + 1 
    print(len(words_dict.items()))
    for word, count in words_dict.items():
        print(str(count), end=" ")

  • jakub_szymanski1
     + 0 comments
    from collections import Counter

n = int(input())
inp = []
for _ in range(n):
    inp.append(input())

inp_set = set(inp)
print(len(inp_set))

c = Counter(inp)
count = [count for item, count in c.items()]
print(*count, sep=' ')

  • apavaloaiestefa1
     + 0 comments

    cool solution using Counter

    from collections import Counter
n = int(input())
words = [input() for _ in range(n)]
c = Counter(words)
print(len(c))
print(*c.values())