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68 changes: 68 additions & 0 deletions alien-dictionary/soobing.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,68 @@
/**
* 문제 설명
* - 주어진 단어들을 활용하여 알파벳 순서를 찾는 문제
*
* 아이디어
* 1) 위상정렬 (👀 다음에 다시풀어보기) - Kahn's Algorithm
* - 어렵다...
*/
function alienOrder(words: string[]): string {
const graph: Map<string, Set<string>> = new Map();
const inDegree: Map<string, number> = new Map(); // 간선의 갯수(첫 시작이 무엇인지 판단하기 위함)

// 단어들에 나오는 모든 문자를 정리 및 초기화
for (const word of words) {
for (const char of word) {
if (!graph.has(char)) {
graph.set(char, new Set());
inDegree.set(char, 0);
}
}
}

// 단어들을 비교해서 알파벳 간의 우선 순서(그래프의 간선) 추출
for (let i = 0; i < words.length - 1; i++) {
const w1 = words[i];
const w2 = words[i + 1];
const minLen = Math.min(w1.length, w2.length);

let foundDiff = false;

for (let j = 0; j < minLen; j++) {
const c1 = w1[j];
const c2 = w2[j];
if (c1 !== c2) {
if (!graph.get(c1)!.has(c2)) {
graph.get(c1)!.add(c2);
inDegree.set(c2, inDegree.get(c2)! + 1);
}
foundDiff = true;
break;
}
}

// 사전순이 아닌 경우 빈문자열 리턴(If the order is invalid, return an empty string.)
if (!foundDiff && w1.length > w2.length) return "";
}

// BFS 위상정렬 시작
const queue: string[] = [];
for (const [char, degree] of inDegree.entries()) {
if (degree === 0) queue.push(char);
}

const result: string[] = [];
while (queue.length > 0) {
const current = queue.shift()!;
result.push(current);

for (const neighbor of graph.get(current)!) {
inDegree.set(neighbor, inDegree.get(neighbor)! - 1);
if (inDegree.get(neighbor) === 0) {
queue.push(neighbor);
}
}
}

return result.length === inDegree.size ? result.join("") : "";
}
Original file line number Diff line number Diff line change
@@ -0,0 +1,43 @@
/**
* 문제 설명
* - preorder(전위순회), inorder(중위순회) 배열을 통해 이진트리를 복원한다
*
*
* 아이디어
* 1) preorder 배열의 요소는 루트 노드이다, 이를 기준으로 inorder 배열을 좌우로 나눈다.
* 2) 좌우로 나눈 inorder 배열의 길이를 통해 preorder 배열의 좌우 서브트리를 구한다.
* 3) 이를 재귀적으로 반복한다.
*/

class TreeNode {
val: number;
left: TreeNode | null;
right: TreeNode | null;
constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
this.val = val === undefined ? 0 : val;
this.left = left === undefined ? null : left;
this.right = right === undefined ? null : right;
}
}

function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
if (preorder.length === 0 || inorder.length === 0) return null;

const inorderIndexMap = new Map<number, number>();
inorder.forEach((value, index) => inorderIndexMap.set(value, index));

let preorderIndex = 0;
const helper = (left: number, right: number): TreeNode | null => {
if (left > right) return null;
const rootValue = preorder[preorderIndex++];
const root = new TreeNode(rootValue);
const index = inorderIndexMap.get(rootValue)!;

root.left = helper(left, index - 1);
root.right = helper(index + 1, right);

return root;
};

return helper(0, inorder.length - 1);
}
31 changes: 31 additions & 0 deletions longest-palindromic-substring/soobing.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,31 @@
/**
* 문제 설명
* - 주어진 문자열에서 가장긴 palindromic substring을 찾는 문제
*
* 아이디어
* 1) palindrom을 찾는 법(중심 확장법) + 홀수ver, 짝수ver 두 가지 경우를 모두 확인
* - two pointer 기법을 이용하여 확장하면서 가장 긴 palindromic substring을 찾는다.
*/
function longestPalindrome(s: string): string {
let maxLength = 0;
let start = 0;

const expand = (l: number, r: number) => {
while (l >= 0 && r < s.length && s[l] === s[r]) {
const currentLength = r - l + 1;
if (currentLength > maxLength) {
maxLength = currentLength;
start = l;
}
l--;
r++;
}
};

for (let i = 0; i < s.length; i++) {
expand(i, i);
expand(i, i + 1);
}

return s.slice(start, start + maxLength);
}
25 changes: 25 additions & 0 deletions rotate-image/soobing.ts
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행 열 바꿔준 다음 아예 행 자체를 reverse 하셨네요. 양끝 포인트들을 swap 하는 방식으로 행 뒤집기를 했었는데 간결하게 구현된 걸 보고 배우고 갑니다.

Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
/**
* 문제 설명
* - 2차원 배열을 90도 in-place로 회전하기
*
* 아이디어
* 1) 대각선 이동 + 좌우 이동
*
*/
/**
Do not return anything, modify matrix in-place instead.
*/
function rotate(matrix: number[][]): void {
const n = matrix.length;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
const temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}

for (let i = 0; i < n; i++) {
matrix[i].reverse();
}
}
33 changes: 33 additions & 0 deletions subtree-of-another-tree/soobing.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,33 @@
/**
* 문제 설명
* - 두 개의 이진트리 중, subTree가 존재하는지 확인하는 문제
*
* 아이디어
* 1) DFS + isSameTree 체크
*/
class TreeNode {
val: number;
left: TreeNode | null;
right: TreeNode | null;
constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
this.val = val === undefined ? 0 : val;
this.left = left === undefined ? null : left;
this.right = right === undefined ? null : right;
}
}

function isSameTree(tree1: TreeNode | null, tree2: TreeNode | null) {
if ((tree1 && !tree2) || (!tree1 && tree2)) return false;
if (tree1 === null && tree2 === null) return true;
if (tree1?.val !== tree2?.val) return false;
return (
isSameTree(tree1?.left ?? null, tree2?.left ?? null) &&
isSameTree(tree1?.right ?? null, tree2?.right ?? null)
);
}

function isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {
if (!root) return false;
if (isSameTree(root, subRoot)) return true;
return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
}