First tried a few loops, worked but slow so didn't pass:

words = [input() for i in range(0, int(input()))]

wordsset = set(words)

def count(words, searchterm):
    return str(len([i for i in words if i == searchterm]))

print(len(wordsset))

counts = [count(words,i) for i in wordsset]

print(" ".join(counts))

So did a one pass attempt which works and passed:

dict = {}
for i in range(0, int(input())):
    strin = input()
    count = dict.get(strin, 0)
    dict[strin] = count+1
    
print(len(dict))
print(' '.join(map(str, dict.values())))

import collections

n = int(input()) odi = collections.OrderedDict()

for word in range(n): word = input() odi[word] = odi.setdefault(word, 0) + 1

print(len(odi)) print(*list(odi.values()))

Dont crtl c/v ! Read and pls try to comprehend

n = int(input()) # Number of words l = [] # List to store the words seen = {} # Dictionary to keep track of occurrences

for i in range(n): word = input() # Input the word l.append(word) # Append word to list if word in seen: seen[word] += 1 # Increment count if word is already in the dictionary else: seen[word] = 1 # Initialize count to 1 if it's the first occurrence of the word

result = [] for word in l: if seen.get(word) is not None: result.append(seen[word]) # Append the count of occurrences del seen[word] # Remove the word from the dictionary to avoid duplicates in output

print(len(result)) # Print the number of distinct words print(*result) # Print the counts of occurrences in the order of appearance

from collections import defaultdict

dic=defaultdict(int)

for i in range(int(input())):

dic[input()]+=1

print(len(dic))

print(' '.join([ str(dic[x]) for x in dic.keys() ]))