diff --git a/find-minimum-in-rotated-sorted-array/KwonNayeon.py b/find-minimum-in-rotated-sorted-array/KwonNayeon.py
index bef237e1b..fc97881fe 100644
--- a/find-minimum-in-rotated-sorted-array/KwonNayeon.py
+++ b/find-minimum-in-rotated-sorted-array/KwonNayeon.py
@@ -1,10 +1,8 @@
 """
-Constraints:
-- n == nums.length
-- 1 <= n <= 5000
-- -5000 <= nums[i] <= 5000
-- All the integers of nums are unique.
-- nums is sorted and rotated between 1 and n times.
+Input/Output/Constraints:
+- Input: A rotated sorted array (e.g., [4,5,6,7,0,1,2])
+- Output: The minimum element in the array (e.g., 0)
+- Constraints: Algorithm must run in O(log n) time
 
 Time Complexity: O(log n) 
 - 이진 탐색을 사용하므로 매 단계마다 탐색 범위가 절반으로 줄어듦
@@ -16,19 +14,17 @@
 1. 이진 탐색(Binary Search) 활용
 2. mid와 right 값을 비교하여 조건을 나눔
   - Case 1: nums[mid] > nums[right]
-    - 오른쪽 부분이 정렬되어 있지 않음
     - 최솟값은 mid 오른쪽에 존재
     - left = mid + 1
   - Case 2: nums[mid] <= nums[right]
-    - mid부터 right까지는 정렬되어 있음
     - 최솟값은 mid를 포함한 왼쪽에 존재 가능
     - right = mid
+3. Pivot이 일어난 지점의 값을 반환
 """
 
 class Solution:
     def findMin(self, nums: List[int]) -> int:
-        left = 0
-        right = len(nums) - 1
+        left, right = 0, len(nums) - 1
 
         while left < right:
             mid = (left + right) // 2
diff --git a/merge-two-sorted-lists/KwonNayeon.py b/merge-two-sorted-lists/KwonNayeon.py
index 327e16b34..0403be50c 100644
--- a/merge-two-sorted-lists/KwonNayeon.py
+++ b/merge-two-sorted-lists/KwonNayeon.py
@@ -4,6 +4,8 @@
  2. -100 <= Node.val <= 100
  3. list1 and list2 are sorted in non-decreasing order
 
+<Solution 1>
+
 Time Complexity: n과 m이 각각 list1과 list2의 길이를 나타낼 때, O(n + m)
  - 각 노드를 한 번씩만 방문하기 때문
 
@@ -43,3 +45,12 @@ def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) ->
             current.next = list2
 
         return result.next
+
+"""
+<Solution 2>
+Time Complexity:
+
+Space Complexity:
+
+풀이 방법:
+"""