def solution(s):
s = [i.split(',') for i in s[2:-2].split('},{')]
s.sort(key=len)
answer = []
answer.append(int(s[0][0]))
for i in range(1, len(s)) :
k = [x for x in s[i] if x not in s[i-1]]
answer.append(int(k[0]))
return answer
ποΈ λ¦¬μ€νΈ μμμ κΈΈμ΄μ λ°λΌμ μ λ ¬
π§© counter
+ regular language
λ‘ λμ± κ°λ¨νκ² νμ΄ν μ μλ€.
import re
from collections import Counter
s="{{1,2,3},{2,1},{1,2,4,3},{2}}"
print(Counter(re.findall('\d+', s)))
# μΆλ ₯
# Counter({'2': 4, '1': 3, '3': 2, '4': 1})