Merge pull request #2020 from 1lsang/main

[1lsang] WEEK 01 Solutions

Author: dalestudy[bot]

contains-duplicate/1lsang.ts

@@ -0,0 +1,5 @@
+function containsDuplicate(nums: number[]): boolean {
+    // Javascript의 Set은 내부적으로 해시테이블을 이용해서, O(n)으로 중복 제거
+    const uniqueNums = Array.from(new Set(nums).values());
+    return uniqueNums.length !== nums.length;
+};

top-k-frequent-elements/1lsang.ts

@@ -0,0 +1,15 @@
+// Runtime: 9ms / Memory: 60MB
+function topKFrequent(nums: number[], k: number): number[] {
+    const numFrequency:Record<number, number> = {}; // { n: freq }
+
+    for (let i=0; i<nums.length; i++) {
+        const n = nums[i];
+        if (numFrequency[n]) numFrequency[n] = numFrequency[n] + 1;
+        else numFrequency[n] = 1;
+    }
+
+    return Object.entries(numFrequency)
+        .sort((a, b) => b[1] - a[1])
+        .slice(0, k)
+        .map(([_, n])=> n);
+};

two-sum/1lsang.ts

@@ -0,0 +1,28 @@
+// 1. 가장 간단한 방법 이중 for문 (Runtime: 175ms / Memory: 55.8MB)
+// function twoSum(nums: number[], target: number): number[] {
+//     const length = nums.length;
+//     for (let i = 0; i < length; i++) {
+//         for (let j = 0; j < length; j++) {
+//             if (i === j) continue;
+//             if (nums[i] + nums[j] === target) return [i, j]
+//         }
+//     }
+//     return [];
+// };
+
+// 2. 시간 복잡도 개선 (Runtime: 9ms / Memory: 60.4MB)
+function twoSum(nums: number[], target: number): number[] {
+    const numObj = {};
+    nums.forEach((num, index) => {numObj[num] ? numObj[num].push(index) : numObj[num]=[index]});
+    for (let i = 0; i<nums.length; i++) {
+        const num = target - nums[i];
+        if (numObj[num]?.length === 1) {
+            if (i === numObj[num][0]) continue;
+            return [i, numObj[num][0]];
+        }
+        else if (numObj[num]?.length === 2) {
+            return numObj[num];
+        }
+    }
+    return [];
+};