카카오 신입 공채 1차 코딩 테스트 문제 5. 뉴스 클러스터링(난이도: 중) (Python)

2017/11/03 - [공부] - 카카오 신입 공채 1차 코딩 테스트 문제 5. 뉴스 클러스터링(난이도: 중)

Java에 이어 Python으로도 작성해 봤다.

import re
 
 
def str_split_by_2(in_str):
    str_list = []
 
    for i in range(0, len(in_str) - 1):
        temp_str = in_str[i:i + 2]
        pattern = re.compile('[a-zA-Z]{2}')
        if pattern.match(temp_str):
            str_list.append(temp_str)
    str_list.sort()
    return str_list
 
 
def question5(str1, str2):
    if len(str1) < 2 or len(str1) > 1000:
        return None
    if len(str2) < 2 or len(str2) > 1000:
        return None
 
    str1_list = str_split_by_2(str1)
    str2_list = str_split_by_2(str2)
 
    intersection = []
    union = []
 
    index1 = 0
    index2 = 0
 
    # get union
    while True:
        if index1 >= len(str1_list) and index2 >= len(str2_list):
            break
 
        if index1 >= len(str1_list):
            union.append(str2_list[index2])
            index2 += 1
            continue
 
        if index2 >= len(str2_list):
            union.append(str1_list[index1])
            index1 += 1
            continue
 
        if str1_list[index1].lower() == str2_list[index2].lower():
            union.append(str1_list[index1])
            index1 += 1
            index2 += 1
        elif str1_list[index1].lower() < str2_list[index2].lower():
            union.append(str1_list[index1])
            index1 += 1
        elif str1_list[index1].lower() > str2_list[index2].lower():
            union.append(str2_list[index2])
            index2 += 1
 
    # get intersection
    index1 = 0
    index2 = 0
    while True:
        if index1 >= len(str1_list) or index2 >= len(str2_list):
            break
 
        if str1_list[index1].lower() == str2_list[index2].lower():
            intersection.append(str2_list[index1])
            index1 += 1
            index2 += 1
        elif str1_list[index1].lower() < str2_list[index2].lower():
            index1 += 1
        elif str1_list[index1].lower() > str2_list[index2].lower():
            index2 += 1
 
    if len(intersection) == 0 and len(union) == 0:
        print("intersection %d, union %d" % (len(intersection), len(union)))
        return 65536
    elif len(intersection) == 0:
        print("intersection %d, union %d" % (len(intersection), len(union)))
        return 0
    elif len(union) == 0:
        print("intersection %d, union %d" % (len(intersection), len(union)))
        return 65536
 
    return int((len(intersection) / len(union)) * 65536)