From fabb21c6ee1d2e28c2108160610a668d7aea7836 Mon Sep 17 00:00:00 2001
From: wazedkhan <wazedkhan119399@gmail.com>
Date: Thu, 4 Dec 2025 08:22:38 +0600
Subject: [PATCH] LeetCode 11: Container with most water

---
 .../medium/container_with_most_water_11.py    | 32 +++++++++++++++
 tests/test_leetcode.py                        | 40 +++++++------------
 2 files changed, 47 insertions(+), 25 deletions(-)
 create mode 100644 LeetCode/medium/container_with_most_water_11.py

diff --git a/LeetCode/medium/container_with_most_water_11.py b/LeetCode/medium/container_with_most_water_11.py
new file mode 100644
index 0000000..9cd2fb2
--- /dev/null
+++ b/LeetCode/medium/container_with_most_water_11.py
@@ -0,0 +1,32 @@
+# LeetCode 11. Container With Most Water
+from typing import List
+
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        """
+        Pseudocode:
+        The problem is we dont know the max area where we we can store most of the water
+        So we can store from the start to end of the container but if the starting height of container is 1 then
+        max we can store height of 1 as more then that will spile the water regardless the end height
+        so if len is 9 and height is 1 then container can contain Area = length × width
+
+        So, what we can do it use two pointer solution to find the max area
+        - initialize two pointer left and right, left with value of index 0 and
+          right index len(arr)-1 and res which contain most largest value
+        - we will continue until our left and right point cross their path
+        - we will be moving our smallest pointer as we know smallest length contain less water
+        - each time we will check which one is largest previous area which stored in res
+        - at the end we return the res as res will be containing the largest area value we got
+        """
+        res = 0
+        left, right = 0, 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
diff --git a/tests/test_leetcode.py b/tests/test_leetcode.py
index 66698d2..8ec6cb1 100644
--- a/tests/test_leetcode.py
+++ b/tests/test_leetcode.py
@@ -469,28 +469,18 @@ def test_min_number_operations(target, expected):
     assert minOperations(target) == expected
 
 
-# @pytest.mark.performance
-# def test_bank_performance():
-#     from LeetCode.medium.simple_bank_system_2043 import Bank
-
-#     n = 100_000
-#     bank = Bank([1000] * n)  # 100k accounts, each with balance 1000
-#     operations = 200_000
-
-#     start = time.perf_counter()
-
-#     for _ in range(operations):
-#         op_type = random.choice(["deposit", "withdraw", "transfer"])
-#         a1 = random.randint(1, n)
-#         money = random.randint(1, 100)
-
-#         if op_type == "deposit":
-#             bank.deposit(a1, money)
-#         elif op_type == "withdraw":
-#             bank.withdraw(a1, money)
-#         else:  # transfer
-#             a2 = random.randint(1, n)
-#             bank.transfer(a1, a2, money)
-
-#     end = time.perf_counter()
-#     print(f"\nTotal time for {operations} ops on {n} accounts: {end - start:.3f}s")
+# LeetCode 11: Container with most water
+@pytest.mark.parametrize(
+    "height, expected",
+    [
+        ([1, 8, 6, 2, 5, 4, 8, 3, 7], 49),
+        ([1, 1], 1),
+        ([4, 3, 2, 1, 4], 16),
+        ([1, 2, 1], 2),
+    ],
+)
+def test_max_area(height, expected):
+    from LeetCode.medium.container_with_most_water_11 import Solution
+
+    solution = Solution()
+    assert solution.maxArea(height) == expected