[Programmers] 프로그래머스 뉴스 클러스터링 Python

https://school.programmers.co.kr/learn/courses/30/lessons/17677

프로그래머스

 

from collections import Counter


def solution(str1, str2):
    # J(A, B) -> 교집합 크기 / 합집합 크기
    # A와 B가 공집합이면 1

    str1_list = []
    str2_list = []

    for i in range(len(str1) - 1):
        temp = str1[i] + str1[i + 1]
        if temp.isalpha():
            str1_list.append(temp.lower())

    for i in range(len(str2) - 1):
        temp = str2[i] + str2[i + 1]
        if temp.isalpha():
            str2_list.append(temp.lower())


    # 요소가 몇개씩 있는지
    element = set(str1_list + str2_list)
    str1_counter = dict(Counter(str1_list))
    str2_counter = dict(Counter(str2_list))

    # 합집합, 교집합 찾기
    uni = 0
    inter = 0

    for e in element:
        if e in str1_counter.keys() and e in str2_counter.keys():
            max_val = max(str1_counter[e], str2_counter[e])
            min_val = min(str1_counter[e], str2_counter[e])
            uni += max_val
            inter += min_val
        elif e in str1_counter.keys():
            max_val = str1_counter[e]
            min_val = 0
            uni += max_val
            inter += min_val
        elif e in str2_counter.keys():
            max_val = str2_counter[e]
            min_val = 0
            uni += max_val
            inter += min_val

    if uni == 0 and inter == 0:
        return 65536
    else:
        return int(inter / uni * 65536)