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Commit cd569a0

Browse files
Merge branch 'main' into v0.4.0
2 parents 008a3ca + 5e9192e commit cd569a0

12 files changed

+383
-17
lines changed

‎README.md

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‎level-2/2-x-n-타일링.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//완벽한 정답이 아닙니다.
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//정답 1 - jaewon1676
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function solution(n) {
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let dp = [0, 1, 2] // n이 1, 2일때는 바로 답을 출력,
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if (n>2){ // n이 3 이상이면 필요한 만큼의 수 까지만 수를 만들어준다.
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for (let i=3; i<=n; i++){
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dp.push((dp[i-1] + dp[i-2]) % 1000000007);
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}
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}
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return dp[n]
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}
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/*
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n이 1일땐 1, 2일땐 2, 3일땐 3, 4일땐 5 . . 의 식이 보인다.
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n = (n - 1) + (n - 2)의 식으로 구할 수 있고,
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제한 사항을 주의해서 풀어보자. */

‎level-2/3-x-n-타일링.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(n) {
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if (n % 2 !== 0) return 0;
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const getCount = n => {
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const k = n / 2;
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const count = [3, 11, ...Array(k - 2)];
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const divider = 1000000007;
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for (let i = 2; i < k; i++) {
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count[i] = (4 * count[i - 1] - count[i - 2] + divider) % divider;
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}
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return count[count.length - 1];
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};
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return getCount(n);
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}

‎level-2/N-Queen.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//완벽한 정답이 아닙니다.
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//정답 1 - codeisneverodd
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function solution(n) {
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/*
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1. 0번째 행에 0번째 queen을 놓는다.
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2. 그 다음 행의 퀸은 이전 퀸들의 범위와 겹치지 않는 곳에 놓는다. 퀸은 한 행에 반드시 하나 두어야한다.
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3. 마지막 열까지 도달하면 성공으로 간주하고 answer에 1을 더한다.
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4. 0번째 queen의 위치를 바꿔가며 모두 시도한다.
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4. 단, 체스판은 일차원 배열로 선언하고 index = 행, 값 = 열 로 생각한다.
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*/
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let answer = 0;
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const canBePlacedOn = (chess, currentRow) => {
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//해당 행에 둔 queen이 유효한지
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for (let prevRow = 0; prevRow < currentRow; prevRow++) {
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const onDiagonal = currentRow - prevRow === Math.abs(chess[currentRow] - chess[prevRow])
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const onStraight = chess[prevRow] === chess[currentRow]
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if (onDiagonal || onStraight) return false
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}
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return true
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}
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const placeQueen = (chess, currentRow) => {
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//queen을 배치하다가 끝 행에 도착하면 1을 리턴, 도착하지 못하면 0을 리턴하여, 재귀적으로 모든 경우를 합하여 리턴
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let count = 0
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if (currentRow === chess.length) return 1
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for (let currentQueen = 0; currentQueen < n; currentQueen++) {
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//queen을 우선 배치한 후 가능한지 살펴본다.
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chess[currentRow] = currentQueen
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if (canBePlacedOn(chess, currentRow)) count += placeQueen(chess, currentRow + 1)
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}
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return count
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}
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for (let firstQueen = 0; firstQueen < n; firstQueen++) {
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const chess = new Array(n).fill(-1)
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chess[0] = firstQueen
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answer += placeQueen(chess, 1)
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}
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return answer;
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}

‎level-2/교점에-별-만들기.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(line) {
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const getCrossPoint = ([A, B, E], [C, D, F]) => {
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if (A * D - B * C === 0) return [Infinity, Infinity];
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return [(B * F - E * D) / (A * D - B * C), (E * C - A * F) / (A * D - B * C)];
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}; //문제 설명 최하단 참조
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const crossPoints = line.flatMap((lineA, i) =>
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line
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.slice(i + 1)
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.map(lineB => getCrossPoint(lineA, lineB))
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.filter(([x, y]) => Number.isInteger(x) && Number.isInteger(y))
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);
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const generateCanvas = crossPoints => {
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const xPoints = [...crossPoints.map(([x, y]) => x)];
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const yPoints = [...crossPoints.map(([x, y]) => y)];
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const [minX, maxX] = [Math.min(...xPoints), Math.max(...xPoints)];
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const [minY, maxY] = [Math.min(...yPoints), Math.max(...yPoints)];
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const xLength = Math.abs(maxX - minX) + 1;
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const yLength = Math.abs(maxY - minY) + 1;
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return {
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canvas: Array.from({ length: yLength }, () => Array(xLength).fill('.')),
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draw([x, y], value) {
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this.canvas[Math.abs(y - maxY)][Math.abs(x - minX)] = value;
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},
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print() {
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return this.canvas.map(row => row.join(''));
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},
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};
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};
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const canvas = generateCanvas(crossPoints);
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crossPoints.forEach(point => {
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canvas.draw(point, '*');
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});
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return canvas.print();
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}

‎level-2/멀리-뛰기.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(n) {
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if (n < 2) return 1;
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const count = [0, 1, 2, ...Array(n - 2).fill(0)];
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count.forEach((_, i) => {
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if (i > 2) count[i] = (count[i - 2] + count[i - 1]) % 1234567;
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});
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return count[n];
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}
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//재귀를 사용하면 콜스택 오버플로우가 발생합니다.

‎level-2/방문-길이.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(dirs) {
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const move = ([x, y], dir) => {
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let next = [x, y];
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if (dir === 'U') next = [x, y + 1];
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if (dir === 'D') next = [x, y - 1];
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if (dir === 'R') next = [x + 1, y];
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if (dir === 'L') next = [x - 1, y];
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if (Math.abs(next[0]) > 5 || Math.abs(next[1]) > 5) return [x, y];
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return next;
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};
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const isSameRoute = ([s1, e1], [s2, e2]) => {
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const isSamePoint = ([x1, y1], [x2, y2]) => x1 === x2 && y1 === y2;
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return (isSamePoint(s1, s2) && isSamePoint(e1, e2)) || (isSamePoint(s1, e2) && isSamePoint(s2, e1));
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};
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const trace = {
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visited: [],
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visit(start, end) {
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if (start[0] === end[0] && start[1] === end[1]) return;
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if (!this.visited.find(route => isSameRoute(route, [start, end]))) this.visited.push([start, end]);
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},
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};
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let current = [0, 0];
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dirs.split('').forEach(dir => {
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const next = move(current, dir);
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trace.visit(current, next);
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current = next;
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});
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return trace.visited.length;
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}

‎level-2/이진-변환-반복하기.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(s) {
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const removeZero = s => {
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const removed = s
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.split('')
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.filter(n => n !== '0')
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.join('');
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return { removed, count: s.length - removed.length };
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};
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const convertToBinary = (s, turnCount, removedCount) => {
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if (s === '1') return [turnCount, removedCount];
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const { removed, count } = removeZero(s);
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return convertToBinary(removed.length.toString(2), turnCount + 1, removedCount + count);
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};
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return convertToBinary(s, 0, 0);
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}
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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(n, wires) {
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const hasOneOfWire = (tree, [a, b]) => tree.includes(a) || tree.includes(b);
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const convertWiresToTree = wires => [...new Set(wires.flat())];
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const generateTree = (wires, tree) => {
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if (!wires.find(wire => hasOneOfWire(tree, wire))) return tree;
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const nextWires = wires.filter(wire => !hasOneOfWire(tree, wire));
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const nextTree = [...tree, ...convertWiresToTree(wires.filter(wire => hasOneOfWire(tree, wire)))];
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return [...new Set(generateTree(nextWires, nextTree))];
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};
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let minDiff = Infinity;
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const length = convertWiresToTree(wires).length;
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wires.forEach((_, i) => {
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const [initWire, ...remainWires] = wires.filter((_, j) => j !== i);
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const lengthA = generateTree(remainWires, convertWiresToTree([initWire])).length;
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const diff = Math.abs(lengthA - (length - lengthA));
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minDiff = Math.min(diff, minDiff);
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});
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return minDiff;
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}

‎level-2/줄-서는-방법.js

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//https://github.com/codeisneverodd/programmers-coding-test
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//더 좋은 풀이가 존재할 수 있습니다.
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//정답 1 - codeisneverodd
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function solution(n, k) {
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const getFactorial = n => {
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const result = [1, 1, 2, ...Array(n - 2)];
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result.forEach((_, i) => {
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if (i > 2) result[i] = result[i - 1] * i;
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});
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return result;
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};
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const getDivision = (dividend, divisor) => {
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const quotient = Math.floor(dividend / divisor);
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const remainder = dividend % divisor;
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return [quotient, remainder];
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};
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const stepCount = getFactorial(n).reverse();
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const generateSteps = (k, step) => {
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const [q, r] = getDivision(k, stepCount[step]);
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if (r === 0) return [q];
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return [q, ...generateSteps(r, step + 1)];
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};
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const answer = [];
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const steps = generateSteps(k - 1, 0);
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const notUsedNums = Array.from({ length: n }, (_, i) => i + 1);
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steps.slice(1).forEach(q => {
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answer.push(notUsedNums[q]);
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notUsedNums.splice(q, 1);
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});
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return [...answer, ...notUsedNums];
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}

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