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

Word Order

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  • bongoehit
     + 1 comment
    n = int(input())

arr = []
for i in range(n):
    arr.append(input())
    
K = (sorted(set(arr), key = arr.index ))  

count = []

for i in range(0,len(K)):
    
    count.append(0)
    for j in range(0,len(arr)):    
        if(K[i] == arr[j]):
            count[i] = count[i] + 1
            
print(len(K))
print(*count, sep=" ")

  • efremovdti
     + 0 comments
    from collections import Counter

n = int(input())
words_list = [input() for n in range(n)]
words_count = Counter(words_list)

print(len(words_count))
print(*words_count.values())

  • spoudel522
     + 0 comments

    from collections import Counter n=int(input()) l=[] for i in range(0,n): l.append(input()) x=Counter(l) #counter method will return the character and its no of occurance in dictionay print(len(x)) print(*x.values()) #unpaking dictionay values

  • adhiraj_majumda1
     + 0 comments

    Easy to understand solution

    n = int(input())

dictV={}
lists = []
for i in range(n):
    lists.append(input())

for i in lists:
    if i in dictV:
        dictV[i]+=1
    else:
        dictV[i]=1
output = list(dictV.values())
print(len(dictV))
print(' '.join(str(i) for i in output))

  • adhiraj_majumda1
     + 0 comments

    Easy to understand solution

    Enter your code here. Read input from STDIN. Print output to STDOUT

    n = int(input())

    dictV={} lists = [] for i in range(n): lists.append(input())

    for i in lists: if i in dictV: dictV[i]+=1 else: dictV[i]=1 output = list(dictV.values()) print(len(dictV)) print(' '.join(str(i) for i in output))

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