기본을 충실하게

function solution(n, a, b) { let answer = 0; let order = new Array(n) .fill(0) .map((_, index) => (index + 1 === a || index + 1 === b ? 1 : 0)); let temp = []; while (order.length !== 1) { for (let i = 0; i < order.length; i += 2) { if ((order[i] || order[i + 1]) === 0) { temp.push(0); } else if (order[i] === 1 && order[i + 1] === 1) { answer++; return answer; } else { temp.push(1); } } order = ..

//sol1 function solution(n, words) { const answer = [0, 0]; const temp = [words[0]]; let wrongNumber = 0; for (let i = 1; i < words.length; i++) { if ( temp.includes(words[i]) || temp[i - 1]?.slice(temp[i - 1].length - 1) !== words[i]?.slice(0, 1) ) { wrongNumber = i + 1; break; } else { temp.push(words[i]); } } if (wrongNumber === 0) return answer; return [ isInteger(wrongNumber / n) ? n : wron..

function solution(s) { if (!s) return 1; let stack = []; for (let i = 0; i 0 && stack[stack.length - 1] === s[i]) { stack.pop(); } else { stack.push(s[i]); } } return stack.length === 0 ? 1 : 0; }

// sol1 function solution(n) { let answer = 0; const count1 = convertBinary(n); for (let i = n + 1; i

function solution(n) { let previous = 0; let next = 1; let current = 1; for (let i = 1; i

function solution(n) { let result = 0; for (let i = 1; i

// sol1 자바스크립트 내장 함수 사용 최소화 function solution(s) { if (s === "1") return [0, 0]; let binary = s; let [zeroDelCount, changeCount] = [0, 0]; while (binary !== "1") { let temp = ""; for (let i = 0; i < binary.length; i++) { binary[i] === "1" ? (temp += "1") : zeroDelCount++; } binary = ""; let tempLength = temp.length; let quotient = null; while (quotient !== 0) { quotient = Math.floor(tempLength /..

function solution(A,B){ let answer = 0; const initialValue = 0; const [first,second] = [A.sort((a,b) => a-b), B.sort((a,b) => b-a)] answer = first.reduce((prev,curr,i)=> prev + curr * second[i],initialValue) return answer; } 두 배열 원소의 곱의 누적이 최소가 되게하는 문제로 a배열의 가장 큰수와 b배열의 가장 작은수를 곱해서 누적시켜서 해결했다.