Studying programming

Solution in Javascript I const k = 3; const a = [-1, -3, 4, 2]; function angryProfessor(k, a) { const p = a.filter(v => v < 1).length < k ? 'YES' : 'NO' return p; } angryProfessor(k, a); 1. To filter elements which are less than 1, use filter() 2. length() is used to get the number of the elements. for loop 와 if 를 사용하는 것보다 내장 함수들을 사용하자. Solution in Javascript II const k = 4; const a = [-1, -3, -..

Solution in Javascript const i = 20; const j = 23; const k = 6; function beautifulDays(i, j, k) { let beautifulDay = 0; while(i

The solution in Javascript I (Failed in HackerRank) sort 를 이용. 순서대로 정렬해준 후에 동일한 값의 (last index - first index) + 1 (0부터 시작하니까) 해외 블로그 접근이 좋긴했지만 결과적으로 HackerRank에서는 fail. const arr = [1, 2, 3, 4, 5, 4, 3, 2, 1, 3, 4, 1] function migratoryBirds(arr) { let bird = 1; let max = bird; //bird 의 시작이 1이라는 것은 at least one bird id는 존재한다는 의미. 나중에 더 큰 값이오면 대체. let result = 0; arr.sort(); //1,1,1,2,2,3,3,3,4,4..

Solution in JavaScript const n = 5; function utopianTree(n) { let cycle = 1; let height = 1; while (cycle

Example code app.post("/register", (req, res) => { User.findOne({username: req.body.username}, async(err, doc) => { if(err) throw err; if(doc) res.send('User already exist'); if(!doc){ const hashedPassword = await bcrypt.hash(req.body.password, 10); const newUser = new User({ username: req.body.username, password: hashedPassword, }); await newUser.save(); res.send('User Created') } }) }) bcrypt ..

Solution I in Javascript const keyboards = [3, 1]; const drives = [5, 2, 8]; const b = 10; function getMoneySpent(keyboards, drives, b) { let valid = []; keyboards.forEach(k => drives.forEach(d => k + d //1. reduce: arrow function 이용 single value return. //2. max: 조건에 맞게 max value를 리턴. //3. map: 함수이용 each elements 변화시킨다. -> drives와 keyboards each elements 더하기. //4. 합한 값을 budget보다 작거나 같은 값으로 filt..