스걱스걱

import java.util.Collections; import java.util.HashMap; import java.util.Map; import java.util.Scanner; public class Main { public static void main(String[] args) { int input = new Scanner(System.in).nextInt(); result(input); } private static void result(int num) { Map map = new HashMap(); int cnt = 0; while(num > 0) { int digit = num % 10; if(digit == 6 || digit == 9) { cnt++; } else { if(map.c..

import java.util.Scanner; public class Main { public static void main(String[] args) { int input = new Scanner(System.in).nextInt(); main(input); } public static void main(int input) { String str = ""; for(int i = 0; i < input; i++) { for(int j = 1; j

def solution(s): answer = '' dic = { 'zero': '0', 'one': '1', 'two': '2', 'three': '3', 'four': '4', 'five': '5', 'six': '6', 'seven': '7', 'eight': '8', 'nine': '9' } temp = '' for word in s: if word.isdigit(): answer += word else: temp += word if temp in dic: answer += dic[temp] temp = '' return int(answer) 풀긴 풀었으나 역시 다른 사람들의 풀이를 보니 아쉽다. for key, value in nums.items(): s = s.replace(key,value)..

import math def solution(progresses, speeds): days = [] answer = [] for p, s in zip(progresses, speeds): days.append(math.ceil((100 - p) / s)) prev = 0 j = 0 answer.append(1) for i in range(1, len(days)): if days[prev] >= days[i] : answer[j] += 1 else: answer.append(1) prev = i j+=1 return answer

해시 문제라길래 해시로 풀려고 엄청 고민했다. 효율성 3, 4번이 계속 실패해서 5시간 넘게 고민해본 문제. 결국 질문하기에 올라온 설명보고 성공했다. 효율성을 좀 더 고려하면서 풀수 있도록 해봐야겠다. def solution(phone_book): answer = True phone_nums = sorted(phone_book) for i in range(1, len(phone_nums)): prev_num = phone_nums[i-1] current_num = phone_nums[i] if(current_num.startswith(prev_num)): return False return answer

def solution(a, b): sum=0 for i in range(len(a)): sum+=a[i]*b[i] answer = sum return answer 기본적으로 풀었는데 zip()함수를 이용한 방법 찾아보기

def solution(answers): answer = [] p1 = [1,2,3,4,5] p2 = [2,1,2,3,2,4,2,5] p3 = [3,3,1,1,2,2,4,4,5,5] j = 0 k = 0 m = 0 correct = {1: 0, 2: 0, 3: 0} for i in answers: j = 0 if j == len(p1) else j k = 0 if k == len(p2) else k m = 0 if m == len(p3) else m correct[1] += 1 if p1[j] == i else 0 correct[2] += 1 if p2[k] == i else 0 correct[3] += 1 if p3[m] == i else 0 j += 1 k += 1 m += 1 answer = [k ..

정규식 찾아보느라 좀 오래 걸렸다. 이 문제를 풀면서 정규식에 대한 이해를 조금이나마 더 하게 되었다. import re def solution(new_id): answer = '' new_id = new_id.lower() new_id = re.sub('[^a-z0-9-_.]', '', new_id) new_id = re.sub('[.]{1,}', '.', new_id).strip('.') if not new_id : new_id = "a" if len(new_id) >= 16 : new_id = new_id[:15].rstrip('.') if len(new_id)

def solution(board, moves): answer = 0 basket = []; top = -1; for j in moves: for i in range(len(board)): if (board[i][j-1] != 0): current = board[i][j-1] board[i][j-1] = 0 top += 1 basket.append(current) if (top > 0 and basket[top] == basket[top-1]): del basket[top] del basket[top-1] top -= 2 answer += 2 break return answer

def solution(numbers, hand): answer = '' dic = {'L':10, 'R':12} if(hand == "left"): hand = 'L' else: hand = 'R' for n in numbers: if(n in [1,4,7]): answer += 'L' dic['L'] = n elif (n in [3,6,9]): answer += 'R' dic['R'] = n elif (n in [2,5,8,0]): if(n == 0): n = 11 right_distance = abs(n-dic['R'])//3+abs(n-dic['R'])%3 left_distance = abs(n-dic['L'])//3+abs(n-dic['L'])%3 if(right_distance == left_..