diff --git a/coin-change/hi-rachel.py b/coin-change/hi-rachel.py
new file mode 100644
index 000000000..5a927ef26
--- /dev/null
+++ b/coin-change/hi-rachel.py
@@ -0,0 +1,37 @@
+
+# BFS -> 먼저 찾는 해답이 가장 적은 연산/가장 짧은 거리 -> 가장 적은 count부터 계산해 최소한의 동전 수 보장
+# O(amount * n) time, O(amount) space
+
+from collections import deque
+
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        queue = deque([(0, 0)])  # (동전 개수, 현재 금액)
+        visited = set()
+        while queue:
+            count, total = queue.popleft()
+            if total == amount:
+                return count
+            if total in visited:
+                continue
+            visited.add(total)
+            for coin in coins:
+                if total + coin <= amount:
+                    queue.append((count + 1, total + coin))
+        return - 1
+    
+
+# DP 풀이
+# 어떤 금액을 만드는데 필요한 동전 개수를 알면, 그 금액보다 큰 금액을 만드는데 필요한 동전 개수도 알 수 있다.
+# dp[i] = min(dp[i], dp[i - coin] + 1)
+# O(amount * n) time, O(amount) space
+
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [0] + [amount + 1] * amount
+
+        for coin in coins:
+            for i in range(coin, amount + 1):
+                dp[i] = min(dp[i], dp[i - coin] + 1)
+
+        return dp[amount] if dp[amount] < amount + 1 else -1
diff --git a/coin-change/hi-rachel.ts b/coin-change/hi-rachel.ts
new file mode 100644
index 000000000..11dfea8c8
--- /dev/null
+++ b/coin-change/hi-rachel.ts
@@ -0,0 +1,11 @@
+function coinChange(coins: number[], amount: number): number {
+  const dp = [0, ...new Array(amount).fill(amount + 1)];
+  for (let i = 0; i <= amount; i++) {
+    for (const coin of coins) {
+      if (coin <= i) {
+        dp[i] = Math.min(dp[i], dp[i - coin] + 1);
+      }
+    }
+  }
+  return dp[amount] < amount + 1 ? dp[amount] : -1;
+}
diff --git a/find-minimum-in-rotated-sorted-array/hi-rachel.py b/find-minimum-in-rotated-sorted-array/hi-rachel.py
new file mode 100644
index 000000000..d55213bc5
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/hi-rachel.py
@@ -0,0 +1,28 @@
+# 회전된 오름차순 배열에서 최소값을 이진 탐색으로 찾아라
+
+# O(N) time, O(1) space -> 시간 복잡도 충족 x
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        for i in range(1, len(nums)):
+            if nums[i - 1] > nums[i]:
+                return nums[i]
+        return nums[0]
+    
+
+class Solution:
+
+
+# 이진 탐색 풀이
+# O(log n) time, O(1) space
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        low, high = 1, len(nums) - 1
+        while low <= high:
+            mid = (low + high) // 2
+            if nums[mid - 1] > nums[mid]:
+                return nums[mid]
+            if nums[0] < nums[mid]:
+                low = mid + 1
+            else:
+                high = mid - 1
+        return nums[0]
diff --git a/find-minimum-in-rotated-sorted-array/hi-rachel.ts b/find-minimum-in-rotated-sorted-array/hi-rachel.ts
new file mode 100644
index 000000000..08f376f12
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/hi-rachel.ts
@@ -0,0 +1,17 @@
+function findMin(nums: number[]): number {
+  let low = 0,
+    high = nums.length - 1;
+
+  while (low <= high) {
+    let mid = Math.floor((low + high) / 2);
+    if (nums[mid - 1] > nums[mid]) {
+      return nums[mid]; // 회전이 일어난 곳
+    }
+    if (nums[0] < nums[mid]) {
+      low = mid + 1; // 왼쪽은 이미 정렬되어 있어, 오른쪽으로 이동
+    } else {
+      high = mid - 1; // 왼쪽으로 이동
+    }
+  }
+  return nums[0];
+}
diff --git a/maximum-depth-of-binary-tree/hi-rachel.py b/maximum-depth-of-binary-tree/hi-rachel.py
new file mode 100644
index 000000000..0dc2e2d8d
--- /dev/null
+++ b/maximum-depth-of-binary-tree/hi-rachel.py
@@ -0,0 +1,44 @@
+# O(n) time, 모든 노드를 한 번씩 방문
+# O(h) for DFS / O(w) for BFS space, DFS는 재귀 호출 깊이 (h = 트리 높이), BFS는 큐에 들어가는 최대 노드 수 (w = 최대 너비)
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+# BFS 풀이
+from collections import deque
+
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        queue = deque([root])
+        depth = 0
+
+        while queue:
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            depth += 1
+        return depth
+
+# DFS 풀이
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
diff --git a/maximum-depth-of-binary-tree/hi-rachel.ts b/maximum-depth-of-binary-tree/hi-rachel.ts
new file mode 100644
index 000000000..85e8940ae
--- /dev/null
+++ b/maximum-depth-of-binary-tree/hi-rachel.ts
@@ -0,0 +1,57 @@
+/**
+ * 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)
+ *     }
+ * }
+ */
+
+// O(n) time, 모든 노드를 한 번씩 방문
+// O(h) for DFS / O(w) for BFS space, DFS는 재귀 호출 깊이 (h = 트리 높이), BFS는 큐에 들어가는 최대 노드 수 (w = 최대 너비)
+
+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;
+  }
+}
+
+// BFS 풀이
+function maxDepth(root: TreeNode | null): number {
+  if (root === null) return 0;
+
+  const queue: TreeNode[] = [root];
+  let depth = 0;
+
+  while (queue.length > 0) {
+    const levelSize = queue.length;
+
+    for (let i = 0; i < levelSize; i++) {
+      const node = queue.shift();
+      if (node) {
+        if (node.left) queue.push(node.left);
+        if (node.right) queue.push(node.right);
+      }
+    }
+    depth++;
+  }
+  return depth;
+}
+
+// DFS 풀이
+// function maxDepth(root: TreeNode | null): number {
+//   if (root === null) return 0;
+//   const left = maxDepth(root.left);
+//   const right = maxDepth(root.right);
+//   return Math.max(left, right) + 1;
+// }
diff --git a/merge-two-sorted-lists/hi-rachel.py b/merge-two-sorted-lists/hi-rachel.py
new file mode 100644
index 000000000..feac43c7a
--- /dev/null
+++ b/merge-two-sorted-lists/hi-rachel.py
@@ -0,0 +1,33 @@
+# O(n + m) time, n = list1의 노드 수 / m = list2의 노드 수
+# O(1) space, 연결된 새 리스트는 기존 노드들을 재사용해서 만듬
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        dummy = ListNode()
+        tail = dummy
+
+        while list1 and list2:
+            if list1.val < list2.val:
+                tail.next = list1
+                list1 = list1.next
+            else:
+                tail.next = list2
+                list2 = list2.next
+            tail = tail.next
+
+        if list1:
+            tail.next = list1
+        if list2:
+            tail.next = list2
+        return dummy.next
diff --git a/merge-two-sorted-lists/hi-rachel.ts b/merge-two-sorted-lists/hi-rachel.ts
new file mode 100644
index 000000000..91fab1bc6
--- /dev/null
+++ b/merge-two-sorted-lists/hi-rachel.ts
@@ -0,0 +1,45 @@
+/**
+ * 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)
+ *     }
+ * }
+ */
+
+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;
+  }
+}
+
+function mergeTwoLists(
+  list1: ListNode | null,
+  list2: ListNode | null
+): ListNode | null {
+  const dummy = new ListNode();
+  let tail = dummy;
+  while (list1 !== null && list2 !== null) {
+    if (list1.val < list2.val) {
+      tail.next = list1;
+      list1 = list1.next;
+    } else {
+      tail.next = list2;
+      list2 = list2.next;
+    }
+    tail = tail.next;
+  }
+  if (list1 !== null) {
+    tail.next = list1;
+  }
+  if (list2 !== null) {
+    tail.next = list2;
+  }
+  return dummy.next;
+}
diff --git a/word-search/hi-rachel.py b/word-search/hi-rachel.py
new file mode 100644
index 000000000..446cd033f
--- /dev/null
+++ b/word-search/hi-rachel.py
@@ -0,0 +1,33 @@
+# DFS 풀이
+# board 높이: m, 넓이: n, w: 단어의 길이
+# TC: (m * n * 4^w) -> 4^w 상하좌우 탐색 * 글자판 내의 모든 좌표를 상대로 수행함
+# SC: O(m * n + w) -> traversing 집합 메모리는 글자판의 크기의 비례 + dfs 호출 스택의 깊이가 단어의 길이에 비례해 증가
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        n_rows, n_cols = len(board), len(board[0])
+        traversing = set()
+
+        def dfs(row, col, idx):
+            if idx == len(word):
+                return True
+            if not (0 <= row < n_rows and 0 <= col < n_cols):
+                return False
+            if (row, col) in traversing:
+                return False
+            if board[row][col] != word[idx]:
+                return False
+
+            traversing.add((row, col))
+            # 상하좌우 탐색
+            for (r, c) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
+                if dfs(row + r, col + c, idx + 1):
+                    return True
+            traversing.remove((row, col))
+            return False
+
+        for i in range(n_rows):
+            for j in range(n_cols):
+                if dfs(i, j, 0):
+                    return True
+        return False