diff --git a/work/Easy/Two Pointers/392.is-subsequence.py b/work/Easy/Two Pointers/392.is-subsequence.py
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+++ b/work/Easy/Two Pointers/392.is-subsequence.py	
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+"""
+392. Is Subsequence
+Difficulty: Easy
+https://leetcode.com/problems/is-subsequence/
+
+──────────────────────────────────────────────────
+
+Given two strings s and t, return true if s is a subsequence of t, or
+false otherwise.
+
+A subsequence of a string is a new string that is formed from the
+original string by deleting some (can be none) of the characters
+without disturbing the relative positions of the remaining characters.
+(i.e., "ace" is a subsequence of "abcde" while "aec" is not).
+
+
+
+Example 1:
+
+Input: s = "abc", t = "ahbgdc"
+Output: true
+
+Example 2:
+
+Input: s = "axc", t = "ahbgdc"
+Output: false
+
+
+
+Constraints:
+
+        • 0 <= s.length <= 100
+
+        • 0 <= t.length <= 10^4
+
+        • s and t consist only of lowercase English letters.
+
+
+
+Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk
+where k >= 10^9, and you want to check one by one to see if t has its
+subsequence. In this scenario, how would you change your code?
+"""
+
+
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        if len(s) > len(t):
+            return False
+
+        i, j = 0, 0
+        while i < len(s) and j < len(t):
+            if s[i] == t[j]:
+                i += 1
+            j += 1
+
+        return i == len(s)
diff --git a/work/Medium/Array/11.container-with-most-water.py b/work/Medium/Array/11.container-with-most-water.py
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+++ b/work/Medium/Array/11.container-with-most-water.py
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+"""
+11. Container With Most Water
+Difficulty: Medium
+https://leetcode.com/problems/container-with-most-water/
+
+──────────────────────────────────────────────────
+
+You are given an integer array height of length n. There are n
+vertical lines drawn such that the two endpoints of the i^th line are
+(i, 0) and (i, height[i]).
+
+Find two lines that together with the x-axis form a container, such
+that the container contains the most water.
+
+Return the maximum amount of water a container can store.
+
+Notice that you may not slant the container.
+
+
+
+Example 1:
+
+Input: height = [1,8,6,2,5,4,8,3,7]
+Output: 49
+Explanation: The above vertical lines are represented by array
+[1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue
+section) the container can contain is 49.
+
+Example 2:
+
+Input: height = [1,1]
+Output: 1
+
+
+
+Constraints:
+
+        • n == height.length
+
+        • 2 <= n <= 10^5
+
+        • 0 <= height[i] <= 10^4
+"""
+
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:  # noqa: F821
+        res = 0
+        left = 0
+        right = len(height) - 1
+
+        while left < right:
+            area = (right - left) * min(height[left], height[right])
+            res = max(res, area)
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+
+        return res