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7 changes: 7 additions & 0 deletions contains-duplicate/sooooo-an.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,7 @@
function containsDuplicate(nums: number[]): boolean {
return new Set(nums).size !== nums.length;
}
/**
* runtime: 24ms / Beats 27.12%
* memory 74.42MB / Beats 55.98%
*/
24 changes: 24 additions & 0 deletions longest-consecutive-sequence/sooooo-an.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
function longestConsecutive(nums: number[]): number {
if (!nums.length) return 0;

nums.sort((a, b) => a - b);
let longest = 1;
let currentStreak = 1;

for (let i = 1; i < nums.length; i++) {
if (nums[i] === nums[i - 1]) continue;
if (nums[i] === nums[i - 1] + 1) {
currentStreak++;
} else {
longest = Math.max(longest, currentStreak);
currentStreak = 1;
}
}

return Math.max(longest, currentStreak);
}

/**
* * Runtime: 37ms / 74.15%
* * Memory: 67.88MB / 93.54%
*/
16 changes: 16 additions & 0 deletions top-k-frequent-elements/sooooo-an.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
function topKFrequent(nums: number[], k: number): number[] {
const frequentMap = new Map<number, number>();

for (const num of nums) {
frequentMap.set(num, (frequentMap.get(num) ?? 0) + 1);
}

const sorted = Array.from(frequentMap.entries()).sort((a, b) => b[1] - a[1]);
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저는 전체 정렬을 하게되면 더 오래걸릴 거라고 생각해서 힙 구조를 사용해봤습니다. 이 코드를 보고 찾아보니 모든 상황에서 그런 건 아니네요. 배워갑니다!


return sorted.slice(0, k).map(([key]) => Number(key));
}

/**
* Runtime 9ms / 78.90%
* Memory 60.130MB / 66.67%
*/
17 changes: 17 additions & 0 deletions two-sum/sooooo-an.ts
Original file line number Diff line number Diff line change
@@ -0,0 +1,17 @@
function twoSum(nums: number[], target: number): number[] {
const visited: { [key: string]: boolean } = {};
for (let i = 0; i < nums.length; i++) {
const complement = target - nums[i];
if (visited[complement]) {
const complementIdx = nums.indexOf(complement);
return [i, complementIdx];
}
visited[nums[i]] = true;
}
return [];
}

/**
* runtime: O(n)
* memory O(n)
*/