From a18a6164c1fdbfc98449776b1293cfc0ad128e4e Mon Sep 17 00:00:00 2001
From: YeomChaeeun <yce0222@gmail.com>
Date: Mon, 30 Dec 2024 17:40:37 +0900
Subject: [PATCH 1/2] feat: missing-number, merge-two-sorted-lists solution

---
 merge-two-sorted-lists/YeomChaeeun.ts | 29 +++++++++++++++++++++++++++
 missing-number/YeomChaeeun.ts         | 20 ++++++++++++++++++
 2 files changed, 49 insertions(+)
 create mode 100644 merge-two-sorted-lists/YeomChaeeun.ts
 create mode 100644 missing-number/YeomChaeeun.ts

diff --git a/merge-two-sorted-lists/YeomChaeeun.ts b/merge-two-sorted-lists/YeomChaeeun.ts
new file mode 100644
index 000000000..f2996122a
--- /dev/null
+++ b/merge-two-sorted-lists/YeomChaeeun.ts
@@ -0,0 +1,29 @@
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     val: number
+ *     next: ListNode | null
+ *     constructor(val?: number, next?: ListNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.next = (next===undefined ? null : next)
+ *     }
+ * }
+ */
+/**
+ * 두개의 리스트 정렬 - 재귀 알고리즘으로 접근
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(n+m) - 모든 노드를 한 번씩 들르기 때문
+ * - 공간 복잡도: O(n+m) - 함수 호출 스택이 재귀 호출로 인해 사용하기 때문
+ * @param list1
+ * @param list2
+ */
+function mergeTwoLists(list1: ListNode | null, list2: ListNode | null): ListNode | null {
+    if(!(list1 && list2)) return list1 || list2
+    if(list1.val < list2.val) {
+        list1.next = mergeTwoLists(list1.next, list2);
+        return list1
+    } else {
+        list2.next = mergeTwoLists(list2.next, list1);
+        return list2
+    }
+}
diff --git a/missing-number/YeomChaeeun.ts b/missing-number/YeomChaeeun.ts
new file mode 100644
index 000000000..f5067f4a7
--- /dev/null
+++ b/missing-number/YeomChaeeun.ts
@@ -0,0 +1,20 @@
+/**
+ * 주어진 배열의 중간에 없는 숫자 찾기
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(nlogn)
+ * - 공간 복잡도: O(1)
+ * @param nums
+ */
+function missingNumber(nums: number[]): number {
+    if(nums.length === 1) {
+        return nums[0] === 0 ? 1 : 0
+    }
+
+    nums.sort((a, b) => a - b)
+
+    for(let i = 0; i < nums.length; i++) {
+        if(nums[0] !== 0) return 0
+        if(nums[i] + 1 !== nums[i + 1])
+            return nums[i] + 1
+    }
+}

From 9e3a817c28675e46becceb2c98f5795cb6c208da Mon Sep 17 00:00:00 2001
From: YeomChaeeun <yce0222@gmail.com>
Date: Thu, 2 Jan 2025 23:47:30 +0900
Subject: [PATCH 2/2] feat: coin-change solution

---
 coin-change/YeomChaeeun.ts | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)
 create mode 100644 coin-change/YeomChaeeun.ts

diff --git a/coin-change/YeomChaeeun.ts b/coin-change/YeomChaeeun.ts
new file mode 100644
index 000000000..bc0b21f2b
--- /dev/null
+++ b/coin-change/YeomChaeeun.ts
@@ -0,0 +1,20 @@
+/**
+ * 동전들로 금액을 만들때 필요한 최소 동전의 개수 찾기
+ * 알고리즘 복잡도
+ * - 시간 복잡도: O(nxm) 동전의 개수 x 만들어야하는 금액의 크기
+ * - 공간 복잡도: O(m) 주어진 금액에 비례함
+ * @param coins
+ * @param amount
+ */
+function coinChange(coins: number[], amount: number): number {
+    const dp = new Array(amount + 1).fill(amount + 1)
+    dp[0] = 0 // 0원은 0개
+
+    for (const coin of coins) {
+        for (let i = coin; i <= amount; i++) {
+            dp[i] = Math.min(dp[i], dp[i - coin] + 1)
+        }
+    }
+
+    return dp[amount] === amount + 1 ? -1 : dp[amount]
+}