아임파인땡큐 아임파인땡큐

const isSubsequence = (s, t) => { //! Edge case if (s.length > t.length) return false; const t_length = t.length; let subsequence = 0; for (let i = 0; i < t_length; i++) { if (s[subsequence] === t[i]) { // ! if it is matching, increment subsequence subsequence++; } } return subsequence === s.length };

/** * @param {string} s * @param {string} t * @return {boolean} */ var isIsomorphic = function(s, t) { //객체에 index 저장하고, 양쪽 index 가 같으면..?? const sObj = {} const tObj ={} const sArr = []; const tArr = []; for(let i = 0;iel.join('')) let answer = sStr.filter((x)=>!tStr.includes(x)) return !answer.length; }; 처음 푼 풀이 var isIsomorphic = function(s, t) { // Base case: for different length of two stri..

function solution(orders, course) { var answer = []; for(let i=0;i=2){ for(let key in obj){ if(obj[key]===max){ addAnswer.push(key) } } } answer.push(...addAnswer) } return answer.sort() } function combi(len,arr){ const answer = []; const temp = Array(len); function DFS(L,index){ if(L===len) answer.push(temp.slice()) else{ for(let i=index;i

function solution(n) { let arr = Array.from(Array(n), (_, index) => Array(index + 1).fill(0)); let lastNum = (n+1)*n/2 let currentNum = 0; let addNum = n; let currentX = 0; let currentY = -1; while(addNum>0){ for(let i=0;i

function solution(number, k) { let stack = []; for(let i=0;i0&&stack[stack.length-1]0){ stack.pop() k--; } stack.push(number[i]) } stack.splice(stack.length-k,k) return stack.join('') }

function solution(n) { const answer = []; //3진수라고 생각할것. 계속 나누고, //나머지는 answer에 추가하고 몫은 계속 가지고 있는다. let quotient = n; while(quotient>0){ let reminder = quotient%3; quotient = Math.floor(quotient/3); if(reminder===0) {answer.push(4) quotient--} else answer.push(reminder) } return answer.reverse().join('') }

function solution(numbers) { //제일 큰 수가 앞에 갈수록 좋아... //즉 내림차순일 수록 좋음 // 하지만 334 등 붙어있는 수들은 꼭 붙여줘야하고.... //하 어렵네.... var answer = ''; numbers = numbers.map((el)=>el.toString()).sort((a,b)=>b+a-(a+b)) answer = numbers.join("") let isZero = answer.split("").filter((el)=>el!=="0") return isZero.length===0?"0":answer } 참고한 풀이

function solution(bridge_length, weight, truck_weights){ //다리건너는데 필요한 시간 -> bridge_length //즉 트럭하나가 건너려면 최소 bridge_length 시간은 있어야함 let bridgeCrossing = [truck_weights[0]]; let bridgeTimes = [0]; let times = 1; let truckIdx = 1; let plus = false; while(bridgeCrossing.length>0){ times++; bridgeTimes = bridgeTimes.map((el)=>el+1) if(bridgeTimes[0]>=bridge_length){ bridgeCrossing.shift() bridgeTimes..