JAVA_Two Sum_LeetCode 1

JAVA_Two Sum_LeetCode 1 Two Sum 풀이 class Solution { public int[] twoSum(int[] nums, int target) { for(int i=0; i< nums.length; i++){ for(int j = i+1; j< nums.length; j++){ if(nums[i] + nums[j] == target){ return new int[]{i,j}; } }; }; return null; } } * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

JAVA_Palindrome Number_LeetCode 9

JAVA_Palindrome Number_LeetCode 9 Palindrome Number 풀이 class Solution { public boolean isPalindrome(int x) { if(x < 0){ return false; } int cnt=x; int cnt2=0; while(cnt>0){ cnt2 = cnt2*10 + cnt%10; cnt = cnt/10; } return cnt2==x; } } * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

JAVA_Roman to Integer_LeetCode 13

JAVA_Roman to Integer_LeetCode 13 Roman to Integer 풀이 class Solution { public int romanToInt(String s) { if (s == null || s.length() == 0){ return -1; } HashMap<Character, Integer> map = new HashMap<Character, Integer>(); map.put('I', 1); map.put('V', 5); map.put('X', 10); map.put('L', 50); map.put('C', 100); map.put('D', 500); map.put('M', 1000); int len = s.length(); int res = map.get(s.charAt(len - 1)); for (int i = len - 2; i >= 0; i--) { if (map.get(s.charAt(i)) >= map.get(s.charAt(i + 1)))

JAVA_Longest Common Prefix_LeetCode 14

JAVA_Longest Common Prefix_LeetCode 14 Longest Common Prefix 풀이 class Solution { public String longestCommonPrefix(String[] strs) { if(strs.length == 0) return ""; String pre=strs[0]; for(int i=1; i<strs.length; i++){ while(strs[i].indexOf(pre) != 0){ pre = pre.substring(0, pre.length()-1); } } return pre; } } * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

JAVA_Valid Parentheses_LeetCode 20

JAVA_Valid Parentheses_LeetCode 20 Valid Parentheses 풀이 class Solution { public boolean isValid(String s) { HashMap map = new HashMap(); map.put('(',')'); map.put('[',']'); map.put('{','}'); Stack<Character> stack = new Stack<>(); for(char c : s.toCharArray()){ if(c == '(' || c == '{' || c == '['){ stack.push(c); }else{ if(stack.isEmpty()){ return false; } if(map.get(stack.pop()).equals(c)){ continue; }else{ return false; } } } return stack.isEmpty(); } } * 출처 Level up your coding skills and qui

JAVA_Merge Two Sorted Lists_LeetCode 21

JAVA_Merge Two Sorted Lists_LeetCode 21 Merge Two Sorted Lists 풀이 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode mergeTwoLists(ListNode list1, ListNode list2) { if (list1 == null || list2 == null) return list1 == null ? list2 : list1; if (list1.val > list2.val) return mergeTwoLists(

JAVA_Remove Duplicates from Sorted Array_LeetCode 26

JAVA_Remove Duplicates from Sorted Array_LeetCode 26 Remove Duplicates from Sorted Array 풀이 class Solution { public int removeDuplicates(int[] nums) { int i = 0; for (int n : nums){ if (i == 0 || n > nums[i-1]){ nums[i++] = n; } } return i; } } * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

Python_HTML Parser - Part 2_HackerRank

Python_HTML Parser - Part 2_HackerRank HTML Parser - Part 2 풀이 from html.parser import HTMLParser import re class MyHTMLParser(HTMLParser): def handle_comment(self,data): if re.search(r"\n",data): print(">>> Multi-line Comment") print(data.rstrip()) else: print(">>> Single-line Comment") print(data.rstrip()) def handle_data(self,data): if data != '\n': print(">>> Data") print(re.sub('\n','',data)) html = "" for i in range(int(input())): html += input().rstrip() html += '\n' parser = MyHTMLParser

Python_Detect HTML Tags, Attributes and Attribute Values_HackerRank

Python_Detect HTML Tags, Attributes and Attribute Values_HackerRank Detect HTML Tags, Attributes and Attribute Values 풀이 from html.parser import HTMLParser class MyParser(HTMLParser): def handle_starttag(self, tag, attrs): print(tag) if attrs: print(*(f"-> {i[0]} > {i[1]}" for i in attrs), sep='\n') n = int(input()) html = ''.join([input() for _ in range(n)]) parser = MyParser() parser.feed(html) parser.close() * 출처 Detect HTML Tags, Attributes and Attribute Values | HackerRank Parse HTML tags,

Python_XML 1 - Find the Score_HackerRank

Python_XML 1 - Find the Score_HackerRank XML 1 - Find the Score 풀이 import sys import xml.etree.ElementTree as etree def get_attr_number(node): total = 0 for child in node.iter(): total += len(child.attrib) return total if __name__ == '__main__': sys.stdin.readline() xml = sys.stdin.read() tree = etree.ElementTree(etree.fromstring(xml)) root = tree.getroot() print(get_attr_number(root)) * 출처 XML 1 - Find the Score | HackerRank Learn about XML parsing in Python. www.hackerrank.com

Python_Validating UID_HackerRank

Python_Validating UID_HackerRank Validating UID 풀이 import re t = r'^(?=(?:[a-z\d]*[A-Z]){2})(?=(?:\D*\d){3})(?:([a-zA-Z\d])(?!.*\1)){10}$' for x in range(int(input())): match = re.match(t, input()) if match: print("Valid") else: print("Invalid") * 출처 Validating UID | HackerRank Validate all the randomly generated user identification numbers according to the constraints. www.hackerrank.com

Python_XML2 - Find the Maximum Depth_HackerRank

Python_XML2 - Find the Maximum Depth_HackerRank XML2 - Find the Maximum Depth 풀이 import xml.etree.ElementTree as etree maxdepth = 0 def depth(elem, level): global maxdepth level += 1 for child in elem: depth(child, level) maxdepth = max(level,maxdepth) return maxdepth if __name__ == '__main__': n = int(input()) xml = "" for i in range(n): xml = xml + input() + "\n" tree = etree.ElementTree(etree.fromstring(xml)) depth(tree.getroot(), -1) print(maxdepth) * 출처 XML2 - Find the Maximum Depth | Hacke

오떡 상계점 - 2인 B SET

2인 B-SET

피자헛테이크아웃 노원점 - 씬 메가더블세트

케미콤보 피자, 뿜뿜불고기 피자

SQL 문제 45 - LeetCode 262. Trips and Users

SQL 문제 45 - LeetCode 262. Trips and Users Trips and Users 풀이 -- ORACLE SELECT request_at AS Day ,ROUND( SUM( CASE WHEN status!='completed' THEN 1 ELSE 0 END )/COUNT(*),2) AS "Cancellation Rate" FROM Trips t WHERE request_at BETWEEN '2013-10-01' AND '2013-10-03' AND client_id IN (SELECT users_id FROM users WHERE banned='No') AND driver_id IN (SELECT users_id FROM users WHERE banned='No') GROUP BY request_at; -- MYSQL select request_at day, round(sum(status <> 'completed')/count(*), 2) 'Cancellati

Python HackerRank 문제 63 - Re.start() & Re.end()

Python HackerRank 문제 63 - Re.start() & Re.end() Re.start() & Re.end() 풀이 import re s = input() k = input() pattern = re.compile(k) r = pattern.search(s) if not r: print("(-1, -1)") while r: print("({0}, {1})".format(r.start(), r.end() - 1)) r = pattern.search(s,r.start() + 1) * 출처 Re.start() & Re.end() | HackerRank Find the indices of the start and end of the substring matched by the group. www.hackerrank.com

Python HackerRank 문제 64 - Validating Roman Numerals

Python HackerRank 문제 64 - Validating Roman Numerals Validating Roman Numerals 풀이 regex_pattern = r"^M{0,3}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$" import re print(str(bool(re.match(regex_pattern, input())))) * 출처 Validating Roman Numerals | HackerRank Use regex to validate Roman numerals. www.hackerrank.com

Python HackerRank 문제 65 - Validating phone numbers

Python HackerRank 문제 65 - Validating phone numbers Validating phone numbers 풀이 import re pattern = re.compile(r'^([789]\d{9})$') for i in range(int(input())): s = input() m = pattern.search(s) if m: print('YES') else: print('NO') * 출처 Validating phone numbers | HackerRank Check whether the given phone number is valid or not. www.hackerrank.com

Python_Validating and Parsing Email Addresses_HackerRank

Python_Validating and Parsing Email Addresses_HackerRank Validating and Parsing Email Addresses 풀이 import re pattern = re.compile(r"<[a-z][a-zA-Z0-9\-\.\_]+\@[a-zA-Z]+\.[a-zA-Z]{1,3}>") for a in range(int(input())): b = input().split() c = pattern.search(b[1]) if c: print(b[0], c.string) * 출처 Validating and Parsing Email Addresses | HackerRank Print valid email addresses according to the constraints. www.hackerrank.com

Python_Hex Color Code_HackerRank

Python_Hex Color Code_HackerRank Hex Color Code 풀이 import re pattern = re.compile(r"[\s:](#(?:[0-9A-Fa-f]{3}){1,2})") n = int(input()) for i in range(n): s = input() m = pattern.findall(s) for i in m: print(i) * 출처 Hex Color Code | HackerRank Validate Hex color codes in CSS. www.hackerrank.com

Python_HTML Parser - Part 1_HackerRank

Python_HTML Parser - Part 1_HackerRank HTML Parser - Part 1 풀이 from html.parser import HTMLParser class MyHTMLParser(HTMLParser): def handle_starttag(self, tag, attrs): print ("Start :", tag) for attr in attrs: print ("-> {} > {}".format(attr[0],attr[1])) def handle_endtag(self, tag): print ("End :", tag) def handle_startendtag(self, tag, attrs): print('Empty :', tag) for attr in attrs: print ("-> {} > {}".format(attr[0],attr[1])) parser = MyHTMLParser() n = int(input()) for _ in range(n): parse

피자애 노원역점 - 페페로니 피자, 떡갈비 피자

네네치킨 중계2동점 - 순살치즈스노윙, 스팸튀김

순살치즈스노윙, 스팸 튀김

Python HackerRank 문제 58 - Detect Floating Point Number

Python HackerRank 문제 - Python HackerRank 문제 58 - Detect Floating Point Number Detect Floating Point Number 풀이 import re n = int(input()) for i in range(n): t = input() print (bool(re.match(r"^[-+]?[0-9]*\.[0-9]+$",t))) * 출처 Detect Floating Point Number | HackerRank Validate a floating point number using the regular expression module for Python. www.hackerrank.com

Python HackerRank 문제 59 - Map and Lambda Function

Python HackerRank 문제 59 - Map and Lambda Function Map and Lambda Function 풀이 cube = lambda x: pow(x,3) def fibonacci(n): l = [] for i in range(n - 1 if n > 2 else (n)): if len(l) < 2: l.append(i) if len(l) >= 2 and n > 2: z = l[-1] + l[-2] l.append(z) return l if __name__ == '__main__': n = int(input()) print(list(map(cube, fibonacci(n)))) * 출처 Map and Lambda Function | HackerRank This challenge covers the basic concept of maps and lambda expressions. www.hackerrank.com

Python HackerRank 문제 60 - Re.split()

Python HackerRank 문제 60 - Re.split() 풀이 regex_pattern = r"[,.]" import re print("\n".join(re.split(regex_pattern, input()))) * 출처 Re.split() | HackerRank Split the string by the pattern occurrence using the re.split() expression. www.hackerrank.com

Python HackerRank 문제 61 - Group(), Groups() & Groupdict()

Python HackerRank 문제 61 - Group(), Groups() & Groupdict() Group(), Groups() & Groupdict() 풀이 import re s = input() m = re.search(r"([a-z0-9A-Z])\1", s) if m: print(m.groups()[0]) else: print("-1") * 출처 Group(), Groups() & Groupdict() | HackerRank Using group(), groups(), and groupdict(), find the subgroup(s) of the match. www.hackerrank.com

Python HackerRank 문제 62 - Re.findall() & Re.finditer()

Python HackerRank 문제 62 - Re.findall() & Re.finditer() Re.findall() & Re.finditer() 풀이 import re s = re.findall('(?<=[^aeiouAEIOU])[aeiouAEIOU]{2,}(?=[^aeiouAEIOU])',input()) print("\n".join(s) if s else -1) * 출처 Re.findall() & Re.finditer() | HackerRank Find all the pattern matches using the expressions re.findall() and re.finditer(). www.hackerrank.com

SQL 문제 44 - LeetCode 393. UTF-8 Validation

SQL 문제 44 - LeetCode 393. UTF-8 Validation UTF-8 Validation 풀이 class Solution { public boolean validUtf8(int[] data) { int bytes = 0; for (int val: data) { if(bytes == 0){ if((val >> 7) == 0b0){ continue; }else if((val >> 5) == 0b110){ bytes=1; }else if((val >> 4) == 0b1110){ bytes=2; }else if((val >> 3) == 0b11110){ bytes=3; }else{ return false; } }else{ if((val >> 6) == 0b10){ bytes--; }else{ return false; } } } return bytes == 0; } } 문제가 어려운건 아닌데,, 클래스명과 메서드명을 반드시 맞춰주자.. 안그러면 'cannot find sym

Python HackerRank 문제 23 - Alphabet Rangoli

Python HackerRank 문제 23 - Alphabet Rangoli 풀이 import string def print_rangoli(size): str = 'abcdefghijklmnopqrstuvwxyz' for i in range(size-1,-size,-1): temp = '-'.join(str[size-1:abs(i):-1]+str[abs(i):size]) print(temp.center(4*size-3,'-')) if __name__ == '__main__': n = int(input()) print_rangoli(n) * 출처 Alphabet Rangoli | HackerRank Let's draw rangoli. www.hackerrank.com

Python HackerRank 문제 24 - Capitalize!

Python HackerRank 문제 24 - Capitalize! 풀이 #!/bin/python3 import math import os import random import re import sys # Complete the solve function below. def solve(s): return ' '.join(map(str.capitalize, s.split(' '))) if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') s = input() result = solve(s) fptr.write(result + '\n') fptr.close() * 출처 Capitalize! | HackerRank Capitalize Each Word. www.hackerrank.com

Python HackerRank 문제 25 - itertools.product()

Python HackerRank 문제 25 - itertools.product() 풀이 from itertools import product a = list(map(int, input().split())) b = list(map(int, input().split())) print(*list(product(a, b))) * 출처 itertools.product() | HackerRank Find the cartesian product of 2 sets. www.hackerrank.com

Python HackerRank 문제 26 - collections.Counter()

Python HackerRank 문제 26 - collections.Counter() collections.Counter() 풀이 from collections import Counter stock = int(input()) sizes = input().split() cnt = Counter(sizes) orders = [input().split() for i in range(int(input()))] bill = 0 for order in orders: if order[0] in cnt.keys() and cnt[order[0]]>0: bill=bill+int(order[1]) cnt[order[0]]-=1 print(bill) * 출처 collections.Counter() | HackerRank Use a counter to sum the amount of money earned by the shoe shop owner. www.hackerrank.com

Python HackerRank 문제 27 - itertools.permutations()

Python HackerRank 문제 26 - itertools.permutations() itertools.permutations()풀이 from itertools import permutations word, num = input().split(" ") permutations = list(permutations(word, int(num))) permutations.sort() for i in permutations: print("".join(i)) * 출처 itertools.permutations() | HackerRank Find all permutations of a given size in a given string. www.hackerrank.com

Python HackerRank 문제 16 - Find a string

Python HackerRank 문제 16 - Find a string Find a string 풀이 def count_substring(string, sub_string): cnt = 0 for i in range(len(string)): if string[i:].startswith(sub_string): cnt += 1 return cnt if __name__ == '__main__': string = input().strip() sub_string = input().strip() count = count_substring(string, sub_string) print(count) * 출처 Find a string | HackerRank Find the number of occurrences of a substring in a string. www.hackerrank.com

Python HackerRank 문제 17 - String Validators

Python HackerRank 문제 17 - String Validators String Validators 풀이 if __name__ == '__main__': s = input() print(any([i.isalnum() for i in s])) print(any([i.isalpha() for i in s])) print(any([i.isdigit() for i in s])) print(any([i.islower() for i in s])) print(any([i.isupper() for i in s])) * 출처 String Validators | HackerRank Identify the presence of alphanumeric characters, alphabetical characters, digits, lowercase and uppercase characters in a string. www.hackerrank.com

Python HackerRank 문제 18 - Text Alignment

Python HackerRank 문제 18 - Text Alignment Text Alignment풀이 thickness = int(input()) #This must be an odd number c = 'H' #Top Cone for i in range(thickness): print((c*i).rjust(thickness-1)+c+(c*i).ljust(thickness-1)) #Top Pillars for i in range(thickness+1): print((c*thickness).center(thickness*2)+(c*thickness).center(thickness*6)) #Middle Belt for i in range((thickness+1)//2): print((c*thickness*5).center(thickness*6)) #Bottom Pillars for i in range(thickness+1): print((c*thickness).center(thickn

Python HackerRank 문제 19 - Designer Door Mat

Python HackerRank 문제 19 - Designer Door Mat Designer Door Mat 풀이 n, m = map(int,input().split()) pattern = [('.|.'*(2*i + 1)).center(m, '-') for i in range(n//2)] print('\n'.join(pattern + ['WELCOME'.center(m, '-')] + pattern[::-1])) * 출처 Designer Door Mat | HackerRank Print a designer door mat. www.hackerrank.com

Python HackerRank 문제 20 - String Split and Join

Python HackerRank 문제 20 - String Split and Join String Split and Join 풀이 def split_and_join(line): # write your code here return '-'.join(line.split(' ')) if __name__ == '__main__': line = input() result = split_and_join(line) print(result) * 출처 String Split and Join | HackerRank Use Python's split and join methods on the input string. www.hackerrank.com

Python HackerRank 문제 21 - Text Wrap

Python HackerRank 문제 21 - Text Wrap 풀이 import textwrap def wrap(string, max_width): return "\n".join(textwrap.wrap(string, max_width)) if __name__ == '__main__': string, max_width = input(), int(input()) result = wrap(string, max_width) print(result) * 출처 Text Wrap | HackerRank Wrap the given text in a fixed width. www.hackerrank.com

Python HackerRank 문제 22 - String Formatting

Python HackerRank 문제 22 - String Formatting String Formatting 풀이 def print_formatted(number): w = len(bin(number))-2 for i in range(1, number+1): print('{0:{width}d} {0:{width}o} {0:{width}X} {0:{width}b}'.format(i, width=w)) if __name__ == '__main__': n = int(input()) print_formatted(n) * 출처 String Formatting | HackerRank Print the formatted decimal, octal, hexadecimal, and binary values for $n$ integers. www.hackerrank.com

Python HackerRank 문제 8 - Find the Runner-Up Score!

Python HackerRank 문제 8 - Find the Runner-Up Score! Find the Runner-Up Score! 풀이 if __name__ == '__main__': n = int(input()) arr = map(int, input().split()) print (sorted(set(arr))[-2]) * 출처 Find the Runner-Up Score! | HackerRank For a given list of numbers, find the second largest number. www.hackerrank.com

Python HackerRank 문제 9 - Nested Lists

Python HackerRank 문제 9 - Nested Lists Nested Lists 풀이 if __name__ == '__main__': students = [] for _ in range(int(input())): name = input() score = float(input()) students.append([name,score]) second_high = sorted(set([i[1] for i in students]))[1] print("\n".join(sorted([i[0] for i in students if i[1] == second_high]))) * 출처 Nested Lists | HackerRank In a classroom of N students, find the student with the second lowest grade. www.hackerrank.com

Python HackerRank 문제 10 - Lists

Python HackerRank 문제 10 - Lists Lists 풀이 if __name__ == '__main__': N = int(input()) li = list() for _ in range(N): command = input().split() try: getattr(li, command[0])(*(map(int, command[1:]))) except AttributeError: exec('{}({})'.format(command[0], 'li')) * 출처 Lists | HackerRank Perform different list operations. www.hackerrank.com Discussion on Lists Challenge Perform different list operations. www.hackerrank.com

Python HackerRank 문제 11 - Finding the percentage

Python HackerRank 문제 11 - Finding the percentage Finding the percentage 풀이 if __name__ == '__main__': n = int(input()) student_marks = {} for _ in range(n): name, *line = input().split() scores = list(map(float, line)) student_marks[name] = scores query_name = input() query_scores = student_marks[query_name] print("{0:.2f}".format(sum(query_scores)/(len(query_scores)))) * 출처 Finding the percentage | HackerRank Store a list of students and marks in a dictionary, and find the average marks obtaine

두툼바삭가츠 상계점 - 통치즈 겹겹이 돈가츠, 가라아게 꼬치

메뉴명 : 통치즈 겹겹이 돈가츠, 가라아게 꼬치

Python HackerRank 문제 12 - sWAP cASE

Python HackerRank 문제 12 - sWAP cASE sWAP cASE 풀이 def swap_case(s): result = [x.lower() if x.isupper() else x.upper() for x in s] return ''.join(result) if __name__ == '__main__': s = input() result = swap_case(s) print(result) * 출처 sWAP cASE | HackerRank Swap the letter cases of a given string. www.hackerrank.com

Python HackerRank 문제 13 - Tuples

Python HackerRank 문제 13 - Tuples Tuples 풀이 if __name__ == '__main__': n = int(raw_input()) integer_list = map(int, raw_input().split()) print(hash(tuple(integer_list))) * 출처 Tuples | HackerRank Learn about tuples and compute hash(T). www.hackerrank.com

Python HackerRank 문제 14 - What's Your Name?

Python HackerRank 문제 14 - What's Your Name? What's Your Name? 풀이 def print_full_name(first, last): print("Hello {} {}! You just delved into python.".format(first,last)) if __name__ == '__main__': first_name = input() last_name = input() print_full_name(first_name, last_name) * 출처 What's Your Name? | HackerRank Python string practice: Print your name in the console. www.hackerrank.com

Python HackerRank 문제 15 - Mutations

Python HackerRank 문제 15 - Mutations Mutations 풀이 def mutate_string(string, position, character): str = list(string) str[position] = character return ''.join(str) if __name__ == '__main__': s = input() i, c = input().split() s_new = mutate_string(s, int(i), c) print(s_new) * 출처 Mutations | HackerRank Understand immutable vs mutable by making changes to a given string. www.hackerrank.com

JAVA 문제 76 - 아무래도이문제는A번난이도인것같다 BAEKJOON 1402

JAVA 문제 76 - 아무래도이문제는A번난이도인것같다 BAEKJOON 1402 아무래도이문제는A번난이도인것같다 풀이 import java.io.*; import java.util.*; public class Main { public static void main (String[] args) throws java.lang.Exception { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int cnt = Integer.parseInt(br.readLine()); for(int i = 0; i < cnt; i++){ StringTokenizer st = new StringTokenizer(br.readLine()); System.out.println("yes"); } } } 문제의 주어진 점은 테스트 케이스, 테스트 케이스의 숫자 2개이다. 앞 숫자는 전체 곱한 값, 뒤 숫자는 전체 더한 값으로 주

JAVA 문제 77 - 24 BAEKJOON 1408

JAVA 문제 77 - 24 BAEKJOON 1408 24 풀이 import java.io.*; import java.util.*; public class Main { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int num = 0; int num2 = 0; StringTokenizer st = new StringTokenizer(br.readLine(), ":"); int h = Integer.parseInt(st.nextToken()); int m = Integer.parseInt(st.nextToken()); int s = Integer.parseInt(st.nextToken()); num = (h * 3600) + (m * 60) + s; st = new StringTokenizer

Python HackerRank 문제 1 - Say &quot;Hello, World!&quot; With Python

Python 문제 1 - Say "Hello, World!" With Python(HackerRank) Say "Hello, World!" With Python 풀이 if __name__ == '__main__': my_string = "Hello, World!"; print my_string print와 변수를 사용할 줄 아는지 확인하는 문제로, "Hello, World!"를 my_string에 담아서 출력하면 끝 * 출처 Say "Hello, World!" With Python | HackerRank Get started with Python by printing to stdout. www.hackerrank.com

Python HackerRank 문제 2 - Python If-Else

Python HackerRank 문제 2 - Python If-Else Python If-Else 풀이 #!/bin/python3 import math import os import random import re import sys if __name__ == '__main__': n = int(input()) if n % 2 != 0: print("Weird") elif n in range(2, 6): print("Not Weird") elif n in range(6, 21): print("Weird") elif n >= 20: print("Not Weird") * 출처 Python If-Else | HackerRank Practice using if-else conditional statements www.hackerrank.com

Python HackerRank 문제 3 - Arithmetic Operators

Python HackerRank 문제 3 - Arithmetic Operators Arithmetic Operators 풀이 if __name__ == '__main__': a = int(input()) b = int(input()) print('{0} \n{1} \n{2}'.format((a + b), (a - b), (a * b))) * 출처 Arithmetic Operators | HackerRank Addition, subtraction and multiplication. www.hackerrank.com

Python HackerRank 문제 4 - Python: Division

Python HackerRank 문제 4 - Python: Division Python: Division 풀이 if __name__ == '__main__': a = int(input()) b = int(input()) print(a//b) print(a/b) * 출처 Python: Division | HackerRank Division using __future__ module. www.hackerrank.com

Python HackerRank 문제 5 - Loops

Python HackerRank 문제 5 - Loops Loops 풀이 if __name__ == '__main__': n = int(input()) for i in range(0,n): print(i*i) * 출처 Loops | HackerRank Practice using "for" and "while" loops in Python. www.hackerrank.com

Python HackerRank 문제 6 - Print Function

Python HackerRank 문제 6 - Print Function Print Function 풀이 if __name__ == '__main__': n = int(input()) for i in range(1,n+1): print(i,end='') * 출처 Print Function | HackerRank Learn to use print as a function www.hackerrank.com

Python HackerRank 문제 7 - List Comprehensions

Python HackerRank 문제 7 - List Comprehensions List Comprehensions 풀이 if __name__ == '__main__': x = int(input()) y = int(input()) z = int(input()) n = int(input()) print(list([i,j,k] for i in range(x+1) for j in range(y+1) for k in range(z+1) if i+j+k !=n)) * 출처 List Comprehensions | HackerRank You will learn about list comprehensions. www.hackerrank.com

SQL 문제 39 - Nth Highest Salary LeetCode 177

SQL 문제 39 - Nth Highest Salary LeetCode 177 Nth Highest Salary 풀이 -- oracle CREATE FUNCTION getNthHighestSalary(num IN NUMBER) RETURN NUMBER IS sal NUMBER; BEGIN SELECT SALARY INTO sal FROM (SELECT SALARY, DENSE_RANK() OVER (ORDER BY SALARY DESC) AS RANK FROM EMPLOYEE) WHERE RANK = num AND ROWNUM = 1; RETURN sal; END; -- MYSQL CREATE FUNCTION getNthHighestSalary(num INT) RETURNS INT BEGIN RETURN ( SELECT DISTINCT SALARY FROM ( SELECT SALARY , DENSE_RANK() OVER (ORDER BY SALARY DESC) AS D_RANK FR

SQL 문제 40 - Consecutive Numbers LeetCode 180

SQL 문제 40 - Consecutive Numbers LeetCode 180 Consecutive Numbers 풀이 SELECT DISTINCT T.NUM AS ConsecutiveNums FROM ( SELECT LAG(NUM) OVER(ORDER BY ID ASC) AS P_NUM , NUM , LEAD(NUM) OVER(ORDER BY ID ASC) AS N_NUM FROM LOGS ) T WHERE T.NUM = T.P_NUM AND T.NUM = T.N_NUM; 특정 값 기준, LAG와 LEAD를 통해 이전 / 이후 값을 찾는 풀이다. * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 41 - Department Highest Salary LeetCode 184

SQL 문제 41 - Department Highest Salary LeetCode 184 Department Highest Salary 풀이 -- ORACLE SELECT T.Department, T.EMPLOYEE, T.SALARY FROM ( SELECT D.NAME AS Department , E.NAME AS EMPLOYEE , E.SALARY AS SALARY , RANK() OVER(PARTITION BY D.ID ORDER BY E.SALARY DESC) RANK FROM EMPLOYEE E INNER JOIN DEPARTMENT D ON E.DEPARTMENTID = D.ID ) T WHERE T.RANK = 1; -- MYSQL, MSSQL SELECT T.Department, T.EMPLOYEE, T.SALARY FROM ( SELECT D.NAME AS Department , E.NAME AS EMPLOYEE , E.SALARY AS SALARY , RANK()

SQL 문제 42 - Exchange Seats LeetCode 626

SQL 문제 42 - Exchange Seats LeetCode 626 Exchange Seats 풀이 -- ORACLE SELECT ID , NVL( CASE WHEN MOD(ID, 2) = 1 THEN LEAD(STUDENT,1) OVER(ORDER BY ID) ELSE LAG(STUDENT,1) OVER(ORDER BY ID) END, STUDENT ) STUDENT FROM SEAT; -- MYSQL SELECT ID , IFNULL( CASE WHEN MOD(ID, 2) = 1 THEN LEAD(STUDENT,1) OVER(ORDER BY ID) ELSE LAG(STUDENT,1) OVER(ORDER BY ID) END, STUDENT ) STUDENT FROM SEAT; -- MSSQL SELECT ID , ISNULL( CASE WHEN ID % 2 = 1 THEN LEAD(STUDENT,1) OVER(ORDER BY ID) ELSE LAG(STUDENT,1) OVER(

SQL 문제 43 - Department Top Three Salaries LeetCode 185

SQL 문제 43 - Department Top Three Salaries LeetCode 185 Department Top Three Salaries 풀이 -- ORACLE, MSSQL SELECT DEPARTMENT , EMPLOYEE , SALARY FROM ( SELECT D.NAME AS DEPARTMENT , E.NAME AS EMPLOYEE , E.SALARY AS SALARY , DENSE_RANK() OVER (PARTITION BY DEPARTMENTID ORDER BY SALARY DESC) AS RANK FROM EMPLOYEE E INNER JOIN DEPARTMENT D ON E.DEPARTMENTID = D.ID ) T WHERE T.RANK < 4 ORDER BY DEPARTMENT, SALARY DESC -- MYSQL SELECT DEPARTMENT , EMPLOYEE , SALARY FROM ( SELECT D.NAME AS DEPARTMENT ,

네네치킨 상계중앙점 - 후라이드

메뉴명 : 후라이드

지코바치킨 노원상계2호점 - 순살양념(매운맛)

메뉴명 : 지코바 치킨 순살 양념(매운맛)

JAVA 문제 74 - 약수 BAEKJOON 1037

JAVA 문제 74 - 약수 BAEKJOON 1037 약수 풀이 import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.StringTokenizer; import java.util.Arrays; public class Main { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int num = Integer.parseInt(br.readLine()); int[] numArr = new int[num]; StringTokenizer st = new StringTokenizer(br.readLine()); for(int i = 0; i < num; i++){

JAVA 문제 75 - 쉽게 푸는 문제 BAEKJOON 1292

JAVA 문제 75 - 쉽게 푸는 문제 BAEKJOON 1292 쉽게 푸는 문제 풀이 import java.io.*; import java.util.*; public class Main { public static void main (String[] args) throws java.lang.Exception { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); ArrayList<Integer> list = new ArrayList<Integer>(); StringTokenizer st = new StringTokenizer(br.readLine()); int a = Integer.parseInt(st.nextToken()); int b = Integer.parseInt(st.nextToken()); int sum=0; for(int i = 0; i < 1000; i++) { for(int j = 0;

SQL 문제 32 - Actors and Directors Who Cooperated At Least Three Times LeetCode 1050

SQL 문제 32 - Actors and Directors Who Cooperated At Least Three Times LeetCode 1050 Actors and Directors Who Cooperated At Least Three Times 풀이 SELECT ACTOR_ID , DIRECTOR_ID FROM ACTORDIRECTOR GROUP BY ACTOR_ID, DIRECTOR_ID HAVING COUNT(TIMESTAMP) >= 3 * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 33 - Classes More Than 5 Students LeetCode 596

SQL 문제 33 - Classes More Than 5 Students LeetCode 596 Classes More Than 5 Students 풀이 SELECT CLASS FROM COURSES GROUP BY CLASS HAVING COUNT(STUDENT) >= 5 * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 34 - Not Boring Movies LeetCode 620

SQL 문제 34 - Not Boring Movies LeetCode 620 Not Boring Movies 풀이 -- ORACLE, MYSQL SELECT ID , MOVIE , DESCRIPTION , RATING FROM CINEMA WHERE MOD(ID,2) = 1 AND DESCRIPTION != 'boring' ORDER BY RATING DESC; -- MSSQL SELECT ID , MOVIE , DESCRIPTION , RATING FROM CINEMA WHERE ID % 2 = 1 AND DESCRIPTION != 'boring' ORDER BY RATING DESC; - MSSQL에서 나머지를 구하는데 MOD함수가 먹히지 않고 '%' 연산자를 통해 나머지를 구한다. * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and ge

SQL 문제 35 - Sales Analysis III LeetCode 1084

SQL 문제 35 - Sales Analysis III LeetCode 1084 Sales Analysis III 풀이 - ORACLE, MSSQL SELECT S.PRODUCT_ID, P.PRODUCT_NAME FROM SALES S INNER JOIN PRODUCT P ON S.PRODUCT_ID = P.PRODUCT_ID GROUP BY S.PRODUCT_ID, P.PRODUCT_NAME HAVING MIN(S.SALE_DATE)>= '2019-01-01' AND MAX(S.SALE_DATE) <= '2019-03-31' - MYSQL SELECT S.PRODUCT_ID , P.PRODUCT_NAME FROM SALES S LEFT OUTER JOIN PRODUCT P ON S.PRODUCT_ID = P.PRODUCT_ID GROUP BY S.PRODUCT_ID HAVING SUM(S.SALE_DATE BETWEEN '2019-01-01' AND '2019-03-31') = C

SQL 문제 36 - Reformat Department Table LeetCode 1179

SQL 문제 36 - Reformat Department Table LeetCode 1179 Reformat Department Table 풀이 - ORACLE SELECT * FROM DEPARTMENT PIVOT( SUM(REVENUE) FOR MONTH IN ( 'Jan' JAN_REVENUE, 'Feb' FEB_REVENUE, 'Mar' MAR_REVENUE, 'Apr' APR_REVENUE, 'May' MAY_REVENUE, 'Jun' JUN_REVENUE, 'Jul' JUL_REVENUE, 'Aug' AUG_REVENUE, 'Sep' SEP_REVENUE, 'Oct' OCT_REVENUE, 'Nov' NOV_REVENUE, 'Dec' DEC_REVENUE ) ); - MYSQL SELECT ID , SUM(IF(MONTH = 'Jan', REVENUE, NULL)) AS JAN_REVENUE , SUM(IF(MONTH = 'Feb', REVENUE, NULL)) AS FE

SQL 문제 37 - Bank Account Summary II LeetCode 1587

SQL 문제 37 - Bank Account Summary II LeetCode 1587 Bank Account Summary II 풀이 SELECT U.NAME , SUM(T.AMOUNT) AS BALANCE FROM USERS U LEFT OUTER JOIN TRANSACTIONS T ON U.ACCOUNT = T.ACCOUNT GROUP BY U.NAME HAVING SUM(T.AMOUNT) > 10000; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 38 - Rank Scores LeetCode 178

SQL 문제 38 - Rank Scores LeetCode 178 Rank Scores 풀이 - ORACLE, MSSQL SELECT SCORE , DENSE_RANK() OVER(ORDER BY SCORE DESC) AS RANK FROM SCORES; - MYSQL, MSSQL SELECT SCORE , DENSE_RANK() OVER(ORDER BY SCORE DESC) AS 'RANK' FROM SCORES; - 중복 점수는 같은 순위로 정한다음 중복 개수만큼 순위가 밀려나는 게 아니라 다음 순위로 정하는 방법인 DENSE_RANK함수를 사용했다. - MSSQL에선 별칭을 적을 때 ''와 상관없이 둘다 됬지만, ORACLE은 없고 MYSQL은 존재해야 실행됨을 확인하였다. * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get pr

서울시 노원구 후라이드 참 잘하는 집 상계점 - 순살 반반

메뉴 : 핫간장, 후라이드 순살 반반

굽네치킨 상계 6.7동 점 - 고추 바사삭 순살

메뉴명 : 고추 바사삭 순살

SQL 문제 30 - Top Travellers LeetCode 1407

SQL 문제 30 - Top Travellers LeetCode 1407 Top Travellers 풀이 -- ORACLE SELECT NAME , NVL(SUM(R.DISTANCE),0) AS TRAVELLED_DISTANCE FROM USERS U LEFT OUTER JOIN RIDES R ON U.ID = R.USER_ID GROUP BY NAME, U.ID ORDER BY TRAVELLED_DISTANCE DESC, U.NAME ASC; -- MYSQL SELECT NAME , IFNULL(SUM(R.DISTANCE),0) AS TRAVELLED_DISTANCE FROM USERS U LEFT OUTER JOIN RIDES R ON U.ID = R.USER_ID GROUP BY NAME, U.ID ORDER BY TRAVELLED_DISTANCE DESC, U.NAME ASC; -- MSSQL SELECT NAME , ISNULL(SUM(R.DISTANCE),0) AS TRA

SQL 문제 31 - Market Analysis I LeetCode 1158

SQL 문제 31 - Market Analysis I LeetCode 1158 Market Analysis I 풀이 -- ORACLE SELECT U.USER_ID AS BUYER_ID , TO_CHAR(U.JOIN_DATE,'YYYY-MM-DD') AS JOIN_DATE , NVL(COUNT(O.ORDER_ID),0) AS ORDERS_IN_2019 FROM ORDERS O RIGHT OUTER JOIN USERS U ON O.BUYER_ID = U.USER_ID AND TO_CHAR(ORDER_DATE,'YYYY') = '2019' GROUP BY U.USER_ID, U.JOIN_DATE ORDER BY U.USER_ID; -- MYSQL SELECT U.USER_ID AS BUYER_ID , U.JOIN_DATE AS JOIN_DATE , IFNULL(COUNT(O.ORDER_ID),0) AS ORDERS_IN_2019 FROM ORDERS O RIGHT OUTER JOIN U

SQL 문제 27 - The Latest Login in 2020 LeetCode 1890

SQL 문제 27 - The Latest Login in 2020 LeetCode 1890 The Latest Login in 2020 풀이 SELECT USER_ID , MAX(TIME_STAMP) AS LAST_STAMP FROM LOGINS WHERE TIME_STAMP LIKE '%2020%' GROUP BY USER_ID * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 28 - Find Total Time Spent by Each Employee LeetCode 1741

SQL 문제 28 - Find Total Time Spent by Each Employee LeetCode 1741 Find Total Time Spent by Each Employee 풀이 -- ORACLE SELECT TO_CHAR(EVENT_DAY,'YYYY-MM-DD') AS DAY , EMP_ID , SUM(OUT_TIME) - SUM(IN_TIME) AS TOTAL_TIME FROM EMPLOYEES GROUP BY EVENT_DAY, EMP_ID ORDER BY EVENT_DAY ASC; -- MYSQL, MSSQL SELECT EVENT_DAY AS DAY , EMP_ID , SUM(OUT_TIME) - SUM(IN_TIME) AS TOTAL_TIME FROM EMPLOYEES GROUP BY EVENT_DAY, EMP_ID ORDER BY EVENT_DAY ASC; * 출처 Level up your coding skills and quickly land a job.

SQL 문제 29 - Capital Gain/Loss LeetCode 1393

SQL 문제 29 - Capital Gain/Loss LeetCode 1393 Capital Gain/Loss 풀이 SELECT STOCK_NAME , SUM( CASE WHEN OPERATION = 'Buy' THEN -PRICE WHEN OPERATION = 'Sell' THEN PRICE ELSE 0 END ) AS CAPITAL_GAIN_LOSS FROM STOCKS GROUP BY STOCK_NAME; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 7 - Duplicate Emails LeetCode 182

SQL 문제 7 - Duplicate Emails LeetCode 182 Duplicate Emails 풀이 SELECT EMAIL FROM PERSON GROUP BY EMAIL HAVING COUNT(EMAIL) > 1; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 10 - Delete Duplicate Emails LeetCode 196

SQL 문제 10 - Delete Duplicate Emails LeetCode 196 Delete Duplicate Emails 풀이 -- ORACLE, MSSQL DELETE FROM PERSON WHERE ID NOT IN (SELECT MIN(ID) FROM PERSON GROUP BY EMAIL ); -- MYSQL DELETE P FROM PERSON P, PERSON P2 WHERE P.Email = P2.Email AND P.Id > P2.Id mysql에선 delete 시 하위 쿼리의 동일 테이블에서 선택할 수 없기 떄문에 셀프조인을 이용하여 삭제한다. * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com 13.2.2 DELETE Stat

SQL 문제 8 - Calculate Special Bonus LeetCode 1873

SQL 문제 8 - Calculate Special Bonus LeetCode 1873 Calculate Special Bonus 풀이 -- ORACLE SELECT EMPLOYEE_ID , CASE WHEN MOD(EMPLOYEE_ID,2) = 0 OR SUBSTR(NAME,0,1) = 'M' THEN 0 ELSE SALARY END AS BONUS FROM EMPLOYEES; -- MYSQL SELECT EMPLOYEE_ID , CASE WHEN EMPLOYEE_ID % 2 = 0 OR LEFT(NAME,1) = 'M' THEN 0 ELSE SALARY END AS BONUS FROM EMPLOYEES; -- MSSQL SELECT EMPLOYEE_ID , CASE WHEN EMPLOYEE_ID % 2 = 0 OR LEFT(NAME,1) = 'M' THEN 0 ELSE SALARY END AS BONUS FROM EMPLOYEES; * 출처 Level up your coding

SQL 문제 9 - Swap Salary LeetCode 627

SQL 문제 9 - Swap Salary LeetCode 627 Swap Salary 풀이 UPDATE SALARY SET SEX = CASE WHEN SEX = 'f' THEN 'm' ELSE 'f' END; SELECT 말고 오로지 UPDATE만 해야하는 문제다. * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 11 - Fix Names in a Table LeetCode 1667

SQL 문제 11 - Fix Names in a Table LeetCode 1667 Fix Names in a Table 풀이 -- ORACLE SELECT USER_ID , INITCAP(NAME) AS NAME FROM USERS ORDER BY USER_ID; -- MYSQL SELECT USER_ID , CONCAT(UPPER(LEFT(NAME,1)), LOWER(SUBSTRING(NAME,2))) AS NAME FROM USERS ORDER BY USER_ID; -- MSSQL SELECT USER_ID , UPPER(LEFT(NAME,1)) + LOWER(SUBSTRING(NAME,2,LEN(NAME))) AS NAME FROM USERS ORDER BY USER_ID; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get p

SQL 문제 12 - Group Sold Products By The Date LeetCode 1484

SQL 문제 12 - Group Sold Products By The Date LeetCode 1484 Group Sold Products By The Date 풀이 -- ORACLE SELECT TO_CHAR(A.SELL_DATE,'YYYY-MM-DD') SELL_DATE , COUNT(DISTINCT A.PRODUCT) NUM_SOLD , LISTAGG(A.PRODUCT,',') WITHIN GROUP(ORDER BY A.PRODUCT) PRODUCTS FROM (SELECT DISTINCT SELL_DATE,PRODUCT FROM ACTIVITIES) A GROUP BY A.SELL_DATE; -- MYSQL SELECT DATE_FORMAT(A.SELL_DATE,'%Y-%m-%d') SELL_DATE , COUNT(DISTINCT A.PRODUCT) NUM_SOLD , GROUP_CONCAT(DISTINCT A.PRODUCT ) PRODUCTS FROM ACTIVITIES A

SQL 문제 13 - Patients With a Condition LeetCode 1527

SQL 문제 13 - Patients With a Condition LeetCode 1527 Patients With a Condition 풀이 -- ORACLE, MYSQL SELECT PATIENT_ID , PATIENT_NAME , CONDITIONS FROM PATIENTS WHERE REGEXP_LIKE(CONDITIONS,'^DIAB1|^* DIAB1'); -- MSSQL SELECT PATIENT_ID , PATIENT_NAME , CONDITIONS FROM PATIENTS WHERE CONDITIONS LIKE 'DIAB1%' OR CONDITIONS LIKE '% DIAB1%'; - 오라클과 MYSQL에선 REGEXP_LIKE를 이용하여 풀었으나,, - MSSQL에선 지원하지 않아서 LIKE 문으로 따로 작성하였다. * 출처 Level up your coding skills and quickly land a job. This is the best place to e

JAVA 문제 71 - 음양 더하기 프로그래머스

JAVA 문제 71 - 음양 더하기 프로그래머스 음양 더하기 풀이 import java.util.stream.IntStream; class Solution { public int solution(int[] absolutes, boolean[] signs) { return IntStream.range(0, absolutes.length).map(i -> absolutes[i] * (signs[i] != false ? 1 : -1)).sum(); } } * 출처 코딩테스트 연습 - 음양 더하기 어떤 정수들이 있습니다. 이 정수들의 절댓값을 차례대로 담은 정수 배열 absolutes와 이 정수들의 부호를 차례대로 담은 불리언 배열 signs가 매개변수로 주어집니다. 실제 정수들의 합을 구하여 return 하도록 solution 함수를 완성해주세요. 제한사항 absolutes의 길이는 1 이상 1,000 이하입니다. absolutes의 모든 수는 각각 1 이상 1,000 이하입니다. sig

JAVA 문제 72 - 내적 프로그래머스

JAVA 문제 72 - 내적 프로그래머스 내적 풀이 import java.util.stream.IntStream; class Solution { public int solution(int[] a, int[] b) { return IntStream.range(0, a.length).map(x -> a[x] * b[x]).sum(); } } * 출처 코딩테스트 연습 - 내적 문제 설명 길이가 같은 두 1차원 정수 배열 a, b가 매개변수로 주어집니다. a와 b의 내적 을 return 하도록 solution 함수를 완성해주세요. 이때, a와 b의 내적은 a[0]*b[0] + a[1]*b[1] + ... + a[n-1]*b[n-1] 입니다. (n은 a, b의 길이) 제한사항 a, b의 길이는 1 이상 1,000 이하입니다. a, b의 모든 수는 -1,000 이상 1,000 이하입니다. 입출력 예 a b result [1,2,3,4] [-3,-1,0,2] 3 [-1,0,1] [1,0,-1]

JAVA 문제 73 - 소수 만들기 프로그래머스

JAVA 문제 73 - 소수 만들기 프로그래머스 소수 만들기 풀이 class Solution { public int solution(int[] nums) { int answer = 0; for(int i=0; i<nums.length-2; i++) { for(int j=i+1; j < nums.length-1; j++) { for(int k=j+1; k < nums.length; k++) { int sum = nums[i] + nums[j] + nums[k]; boolean isPrime = true; for (int l=2; l*l <= sum; l++) { if (sum % l == 0) { isPrime = false; break; } } if(isPrime) answer++; } } } return answer; } } * 출처 코딩테스트 연습 - 소수 만들기 주어진 숫자 중 3개의 수를 더했을 때 소수가 되는 경우의 개수를 구하려고 합니다. 숫자들이 들어있는 배열 nums가

SQL 문제 1 - Big Countries LeetCode 595

SQL 문제 1 - Big Countries LeetCode 595 Big Countries 풀이 SELECT NAME , POPULATION , AREA FROM WORLD WHERE POPULATION >= 25000000 OR AREA >= 3000000; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 2 - Recyclable and Low Fat Products LeetCode 1757

SQL 문제 2 - Recyclable and Low Fat Products LeetCode 1757 Recyclable and Low Fat Products 풀이 SELECT PRODUCT_ID FROM PRODUCTS WHERE LOW_FATS = 'Y' AND RECYCLABLE = 'Y'; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 3 - Find Customer Referee LeetCode 584

SQL 문제 3 - Find Customer Referee LeetCode 584 Find Customer Referee 풀이 SELECT NAME FROM CUSTOMER WHERE REFEREE_ID != 2 OR REFEREE_ID IS NULL; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 4 - Customers Who Never Order LeetCode 595

SQL 문제 4 - Customers Who Never Order LeetCode 595 Customers Who Never Order 풀이 SELECT C.NAME AS CUSTOMERS FROM CUSTOMERS C LEFT OUTER JOIN ORDERS O ON C.ID = O.CUSTOMERID WHERE O.CUSTOMERID IS NULL ORDER BY C.NAME ASC; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 5 - Combine Two Tables LeetCode 175

SQL 문제 5 - Combine Two Tables LeetCode 175 Combine Two Tables 풀이 SELECT P.FIRSTNAME , P.LASTNAME , A.CITY , A.STATE FROM PERSON P LEFT OUTER JOIN ADDRESS A ON P.PERSONID = A.PERSONID; * 출처 Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview. leetcode.com

SQL 문제 6 - Employees Earning More Than Their Managers LeetCode 181

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