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Merge pull request #356 from HC-kang/main
[강희찬] WEEK 2 Solution
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construct-binary-tree-from-preorder-and-inorder-traversal/HC-kang.ts
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// Definition for a binary tree node. | ||
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; | ||
} | ||
} | ||
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// T.C: O(N) | ||
// S.C: O(N^2) - Slice makes n-1, n-2, ..., 1 for n times. So, it's O(N^2). | ||
function buildTree(preorder: number[], inorder: number[]): TreeNode | null { | ||
if (preorder.length === 0 || inorder.length === 0) { | ||
return null; | ||
} | ||
const root = new TreeNode(preorder[0]); | ||
const idx = inorder.indexOf(preorder[0]); | ||
root.left = buildTree(preorder.slice(1, idx + 1), inorder.slice(0, idx)); | ||
root.right = buildTree(preorder.slice(idx + 1), inorder.slice(idx + 1)); | ||
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return root; | ||
} | ||
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// Not using slice. but I think it's not necessary... first solution is more readable. and that's not so bad. | ||
// T.C: O(N) | ||
// S.C: O(N) | ||
function buildTree(preorder: number[], inorder: number[]): TreeNode | null { | ||
// this tree is consist of unique values | ||
const inorderMap = new Map<number, number>(); | ||
for (const [i, val] of inorder.entries()) { | ||
inorderMap.set(val, i); | ||
} | ||
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function helper(preLeft: number, preRight: number, inLeft: number, inRight: number): TreeNode | null { | ||
if (preLeft > preRight) return null; | ||
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const rootValue = preorder[preLeft]; | ||
const root = new TreeNode(rootValue); | ||
const inRootIdx = inorderMap.get(rootValue)!; | ||
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const leftSize = inRootIdx - inLeft; | ||
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root.left = helper(preLeft + 1, preLeft + leftSize, inLeft, inRootIdx - 1); | ||
root.right = helper(preLeft + leftSize + 1, preRight, inRootIdx + 1, inRight); | ||
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return root; | ||
} | ||
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return helper(0, preorder.length - 1, 0, inorder.length - 1); | ||
} |
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// T.C: O(n) | ||
// S.C: O(n) | ||
function countBits(n: number): number[] { | ||
// T.C: O(1) | ||
// S.C: O(1) | ||
function countBit(num: number): number { | ||
num = num - ((num >>> 1) & 0x55555555); | ||
num = (num & 0x33333333) + ((num >>> 2) & 0x33333333); | ||
num = (num + (num >>> 4)) & 0x0f0f0f0f; | ||
num = num + (num >>> 8); | ||
num = num + (num >>> 16); | ||
return num & 0x3f; | ||
} | ||
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return new Array(n + 1).fill(0).map((_, i) => countBit(i)); | ||
} | ||
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// T.C: O(n) | ||
// S.C: O(n) | ||
function countBits(n: number): number[] { | ||
const dp = new Array(n + 1).fill(0); | ||
for (let i = 1; i <= n; i++) { | ||
dp[i] = dp[i >> 1] + (i & 1); | ||
} | ||
return dp; | ||
} |
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// T.C: O(n) | ||
// S.C: O(n) | ||
function numDecodings(s: string): number { | ||
const NUM_OF_ALPHA = 26; | ||
const memo = new Map<number, number>(); | ||
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function dfs(idx: number): number { | ||
if (idx === s.length) { | ||
return 1; | ||
} | ||
if (s[idx] === '0') { | ||
return 0; | ||
} | ||
if (memo.has(idx)) { | ||
return memo.get(idx)!; | ||
} | ||
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let count = dfs(idx + 1); | ||
if ( | ||
idx + 2 <= s.length && // check if idx + 2 is in the range | ||
parseInt(s.slice(idx, idx + 2), 10) <= NUM_OF_ALPHA | ||
) { | ||
count += dfs(idx + 2); | ||
} | ||
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memo.set(idx, count); | ||
return count; | ||
} | ||
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return dfs(0); | ||
} |
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/** | ||
* The most simple way | ||
*/ | ||
function encode(strs: string[]): string { | ||
return strs.join('🏖️'); | ||
}; | ||
function decode(s: string): string[] { | ||
return s.split('🏖️'); | ||
}; | ||
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// T.C: O(n) | ||
// S.C: O(n) | ||
function encode(strs: string[]): string { | ||
return strs.map((s) => s.length + '#' + s).join(''); | ||
} | ||
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// T.C: O(n) | ||
// S.C: O(n) | ||
function decode(s: string): string[] { | ||
const res: string[] = []; | ||
let curIdx = 0; | ||
while (curIdx < s.length) { | ||
const sepIdx = s.indexOf('#', curIdx); | ||
const len = parseInt(s.slice(curIdx, sepIdx), 10); | ||
res.push(s.slice(sepIdx + 1, sepIdx + 1 + len)); | ||
curIdx = sepIdx + 1 + len; | ||
} | ||
return res; | ||
} |
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// T.C: O(n) | ||
// S.C: O(1) | ||
function isAnagram(s: string, t: string): boolean { | ||
if (s.length !== t.length) return false; | ||
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const NUM_OF_ALPHA = 26; | ||
const A_CODE = 'a'.charCodeAt(0); | ||
const bucket = new Array(NUM_OF_ALPHA).fill(0); // S.C: O(1) | ||
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for (let i = 0; i < s.length; i++) { // T.C: O(n) | ||
bucket[s.charCodeAt(i) - A_CODE]++; | ||
bucket[t.charCodeAt(i) - A_CODE]--; | ||
} | ||
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return bucket.every(count => count === 0); | ||
} |