diff --git a/find-minimum-in-rotated-sorted-array/ohkingtaek.py b/find-minimum-in-rotated-sorted-array/ohkingtaek.py
new file mode 100644
index 0000000000..3959159a59
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/ohkingtaek.py
@@ -0,0 +1,8 @@
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        """
+        시간복잡도: O(n)
+        공간복잡도: O(1)
+        왜 이렇게 해도 solve가 되네요..ㅎㅎ
+        """
+        return min(nums)
diff --git a/maximum-depth-of-binary-tree/ohkingtaek.py b/maximum-depth-of-binary-tree/ohkingtaek.py
new file mode 100644
index 0000000000..0d36c04cbd
--- /dev/null
+++ b/maximum-depth-of-binary-tree/ohkingtaek.py
@@ -0,0 +1,25 @@
+# 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 Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        """
+        시간복잡도: O(n)
+        공간복잡도: O(n)
+        이진 트리 탐색하며 뎁스 증가 계속 하며 진행
+        """
+        depth = 0
+        if not root:
+            return 0
+        array = [(root, 1)]
+        while array:
+            node, d = array.pop()
+            depth = max(d, depth)
+            if node.left:
+                array.append((node.left, d + 1))
+            if node.right:
+                array.append((node.right, d + 1))
+        return depth
diff --git a/merge-two-sorted-lists/ohkingtaek.py b/merge-two-sorted-lists/ohkingtaek.py
new file mode 100644
index 0000000000..817cafd54e
--- /dev/null
+++ b/merge-two-sorted-lists/ohkingtaek.py
@@ -0,0 +1,26 @@
+# Definition for singly-linked list.
+# 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]:
+        """
+        시간복잡도: O(n)
+        공간복잡도: O(1)
+        1. 두 리스트를 합쳐서 새로운 리스트 반환
+        2. 두 리스트를 돌면서 작은 값을 새로운 리스트에 추가
+        3. 두 리스트를 모두 순회하면 남은 것을 새로운 리스트에 추가
+        """
+        root = ListNode(None)
+        node = root
+        while list1 and list2:
+            if list1.val < list2.val:
+                node.next = list1
+                list1 = list1.next
+            else:
+                node.next = list2
+                list2 = list2.next
+            node = node.next
+        node.next = list1 or list2
+        return root.next
diff --git a/word-search/ohkingtaek.py b/word-search/ohkingtaek.py
new file mode 100644
index 0000000000..b86e311e9a
--- /dev/null
+++ b/word-search/ohkingtaek.py
@@ -0,0 +1,32 @@
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        """
+        시간 복잡도: O(M * N * 4^L)
+        공간 복잡도: O(M * N)
+        1. 행렬을 순회하며 단어의 첫 번째 문자와 일치하는 위치를 찾기
+        2. 찾은 위치에서 상하좌우로 이동하며 단어의 다음 문자와 일치하는 위치를 찾기
+        3. 단어의 마지막 문자까지 찾으면 True를 반환, 못 찾으면 False
+        """
+        m, n = len(board), len(board[0])
+
+        def dfs(i, j, k):
+            if i < 0 or i >= m or j < 0 or j >= n or board[i][j] != word[k]:
+                return False
+            if k == len(word) - 1:
+                return True
+            t = board[i][j]
+            board[i][j] = "#"
+            dx, dy = (1, -1, 0, 0), (0, 0, 1, -1)
+            ok = False
+            for d in range(4):
+                if dfs(i + dx[d], j + dy[d], k + 1):
+                    ok = True
+                    break
+            board[i][j] = t
+            return ok
+
+        for i in range(m):
+            for j in range(n):
+                if dfs(i, j, 0):
+                    return True
+        return False