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[gitsunmin] Week15 Solutions #610

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Nov 24, 2024
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45 changes: 45 additions & 0 deletions longest-palindromic-substring/gitsunmin.ts
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/**
* https://leetcode.com/problems/longest-palindromic-substring/
* time complexity : O(n)
* space complexity : O(n^2)
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시간, 공간 복잡도를 이렇게 판단하신 이유가 있으실까요?
제 생각에 시간 복잡도는 최대 N * N/2O(n^2) 이고 공간복잡도는 리턴값을 고려하더라도 O(n), 제외한다면 O(1)로 볼 수 있을 것 같아서요!

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이렇게 계산했어요!
• for 루프: 문자열 길이 n에 따라 n번 반복 → O(n).
• 팰린드롬 확장: 각 중심에서 최대 문자열 길이만큼 확장 → O(n).
• 전체 시간 복잡도: O(n) × O(n) = O(n²).

최악의 경우, 모든 중심에서 최대 확장되므로 O(n²)

*/

function findPalindromeAroundCenter(s: string, leftIndex: number, rightIndex: number): [number, number] {
while (leftIndex >= 0 && rightIndex < s.length && s[leftIndex] === s[rightIndex]) {
leftIndex--; // 왼쪽으로 확장
rightIndex++; // 오른쪽으로 확장
}

// 팰린드롬의 시작과 끝 인덱스 반환
return [leftIndex + 1, rightIndex - 1];
}

function longestPalindrome(s: string): string {
if (s.length <= 1) return s;

let longestPalindromeStartIndex = 0;
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중요한 사항은 아닙니다!
변수명이 조금 지나친 것 같습니다.. 읽기 어려워요ㅠㅠ

let longestPalindromeLength = 0;

for (let centerIndex = 0; centerIndex < s.length; centerIndex++) {
// 홀수 길이 팰린드롬 확인
const [oddStart, oddEnd] = findPalindromeAroundCenter(s, centerIndex, centerIndex);
const oddLength = oddEnd - oddStart + 1;

if (oddLength > longestPalindromeLength) {
longestPalindromeStartIndex = oddStart;
longestPalindromeLength = oddLength;
}

// 짝수 길이 팰린드롬 확인
const [evenStart, evenEnd] = findPalindromeAroundCenter(s, centerIndex, centerIndex + 1);
const evenLength = evenEnd - evenStart + 1;

if (evenLength > longestPalindromeLength) {
longestPalindromeStartIndex = evenStart;
longestPalindromeLength = evenLength;
}
}

// 가장 긴 팰린드롬 부분 문자열 반환
return s.substring(longestPalindromeStartIndex, longestPalindromeStartIndex + longestPalindromeLength);
}
29 changes: 29 additions & 0 deletions subtree-of-another-tree/gitsunmin.ts
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/**
* https://leetcode.com/problems/subtree-of-another-tree/
* time complexity : O(n * m)
* space complexity : O(n)
*/

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

function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
if (!p && !q) return true;
if (!p || !q || p.val !== q.val) return false;
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재귀함수의 early return에서 조심할 점이 있다면 단축평가에서 손해보는 점인 것 같습니다. 그리고 if-else가 아니라 if문 단독으로 쓰는 경우, 개인적으로 각 if 문들의 조건이 무조건 독립적이도록 짜면 어떨까 싶습니다.


return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}

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);
}
32 changes: 32 additions & 0 deletions validate-binary-search-tree/gitsunmin.ts
Original file line number Diff line number Diff line change
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/**
* https://leetcode.com/problems/validate-binary-search-tree/
* time complexity : O(n)
* space complexity : O(n)
*/

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

function validate(node: TreeNode | null, lower: number, upper: number): boolean {
if (!node) return true;
const val = node.val;

if (val <= lower || val >= upper) return false;
if (!validate(node.right, val, upper)) return false;
if (!validate(node.left, lower, val)) return false;

return true;
}

function isValidBST(root: TreeNode | null): boolean {

return validate(root, -Infinity, Infinity);
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👍

}