Merge pull request #2010 from juhui-jeong/main

[juhui-jeong] WEEK 01 solutions #219

Author: juhui-jeong

contains-duplicate/juhui-jeong.ts

@@ -0,0 +1,16 @@
+/*
+시간 복잡도: O(n)
+공간 복잡도: O(n)
+*/
+function containsDuplicate(nums: number[]): boolean {
+  const seen = new Set<number>();
+
+  for (const num of nums) {
+    if (seen.has(num)) {
+      return true;
+    }
+    seen.add(num);
+  }
+
+  return false;
+}

top-k-frequent-elements/juhui-jeong.ts

@@ -0,0 +1,26 @@
+/*
+시간 복잡도: O(n²)
+공간 복잡도: O(n)
+*/
+function topKFrequent(nums: number[], k: number): number[] {
+  const map = new Map<number, number>();
+  let numLength: number[] = nums;
+  let mapArray;
+
+  while (numLength.length > 0) {
+    // 현재 기준 숫자
+    const current = numLength[0];
+    // 갯수 확인
+    const elementNum = numLength.filter((n) => n === current);
+    // 확인한 숫자 제거
+    const filteredNum = numLength.filter((n) => n !== current);
+
+    numLength = filteredNum;
+    map.set(current, elementNum.length);
+  }
+  mapArray = [...map];
+  mapArray.sort((a, b) => b[1] - a[1]);
+
+  // 상위 k개 num만 반환
+  return mapArray.slice(0, k).map(([num, _]) => num);
+}

two-sum/juhui-jeong.ts

@@ -0,0 +1,41 @@
+/*
+시간 복잡도: O(n)
+공간 복잡도: O(n)
+*/
+function twoSum(nums: number[], target: number): number[] {
+  const map = new Map<number, number>();
+
+  for (let i = 0; i < nums.length; i++) {
+    const num = nums[i];
+    const diff = target - num;
+
+    if (map.has(diff)) {
+      return [map.get(diff)!, i];
+    }
+
+    map.set(num, i);
+  }
+
+  return [];
+}
+
+/*
+이전 풀이
+시간 복잡도: O(n²)
+공간 복잡도: O(1)
+function twoSum(nums: number[], target: number): number[] {
+  for (let i = 0; i < nums.length; i++) {
+    for (let j = 0; j < nums.length; j++) {
+      // 동일한 index일 때는 안됨.
+      if (i === j) {
+        continue;
+      }
+      if (nums[i] + nums[j] === target) {
+        return [i, j];
+      }
+    }
+    return [];
+  }
+}
+
+*/