diff --git a/container-with-most-water/aa601.py b/container-with-most-water/aa601.py
new file mode 100644
index 000000000..80d670582
--- /dev/null
+++ b/container-with-most-water/aa601.py
@@ -0,0 +1,15 @@
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        right = len(height) - 1
+        left = 0
+        max_size = 0
+        for line in range(len(height)):
+            if left >= right:
+                break
+            cur_size = (right - left) * min(height[left], height[right])
+            if height[left] < height[right]: # 왼쪽이 작으면 left 이동
+                left += 1
+            else:
+                right -= 1
+            max_size = max(max_size, cur_size)
+        return max_size
diff --git a/longest-increasing-subsequence/aa601.py b/longest-increasing-subsequence/aa601.py
new file mode 100644
index 000000000..57a45ce16
--- /dev/null
+++ b/longest-increasing-subsequence/aa601.py
@@ -0,0 +1,18 @@
+"""
+    prv [-1, -1, -1,  2, 2, 3, 5,    5]
+    dp  [ 1,  1,  1,  2, 2, 3, 4,    4]
+    nums[10,  9,  2,  5, 3, 7, 101, 18]
+    nums[2] = 2
+    nums[4] = 3
+    nums[5] = 7
+"""
+class Solution:
+    def lengthOfLIS(self, nums: List[int]) -> int:
+        dp = [1 for _ in range(len(nums))]
+        prv = [-1 for _ in range(len(nums))]
+        for i in range(len(nums)):
+            for j in range(i + 1, len(nums)):
+                if nums[i] < nums[j]:
+                    if dp[i] + 1 > dp[j]: 
+                        dp[j] = dp[i] + 1
+        return max(dp)
diff --git a/valid-parentheses/aa601.py b/valid-parentheses/aa601.py
new file mode 100644
index 000000000..c8aed600d
--- /dev/null
+++ b/valid-parentheses/aa601.py
@@ -0,0 +1,18 @@
+class Solution:
+    def isValid(self, s: str) -> bool:
+        # stack 활용, 짝이 맞으면 pop하기
+        stack = []
+        dic = dict(zip(")}]", "({[")) # close bracket : open bracket 형태로 사전 생성
+        for char in s:
+            if char in "({[": # open bracket인 경우 stack에 추가
+                stack.append(char)
+            elif char in ")}]" : # close bracket이면서
+                if not stack or stack.pop() != dic[char]: 
+                    return False
+
+        # len()함수 시간복잡도 먹음 O(n)
+        # => not stack 사용하면 개선됨
+        if len(stack) == 0: # stack이 비어있다면 모든 짝이 맞아서 pop되었으므로 true 반환 
+            return True
+        else:
+            return False