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문제 유형 한줄 수정 난이도 normal Note 1. 2차 10번 0들을 0으로 만들기와 로직 유사. 차이점 찾기(previous) Code # -*- coding: utf-8 -*- # UTF-8 encoding when using korean # 4차 2번 def solution(s): s = s.lower() answer = "" previous = s[0] counter = 1 for alphabet in s[1:]: if alphabet == previous: counter += 1 else: answer += previous + str(counter) counter = 1 previous = alphabet answer += previous + str(counter) return answer..