From f28be89eb34d83b7497b10d9da664f5408c96909 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Tue, 5 Apr 2022 06:20:21 +0530
Subject: [PATCH] feat(11-container-with-most-water): add py3 solution

---
 0011_container-with-most-water/README.md      | 28 +++++++++++++++++++
 .../python3/solution.py                       | 17 +++++++++++
 2 files changed, 45 insertions(+)
 create mode 100644 0011_container-with-most-water/README.md
 create mode 100644 0011_container-with-most-water/python3/solution.py

diff --git a/0011_container-with-most-water/README.md b/0011_container-with-most-water/README.md
new file mode 100644
index 0000000..19460e2
--- /dev/null
+++ b/0011_container-with-most-water/README.md
@@ -0,0 +1,28 @@
+You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` 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:**
+
+![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
+
+    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 <= 105`
+*   `0 <= height[i] <= 104`
\ No newline at end of file
diff --git a/0011_container-with-most-water/python3/solution.py b/0011_container-with-most-water/python3/solution.py
new file mode 100644
index 0000000..07407a6
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+++ b/0011_container-with-most-water/python3/solution.py
@@ -0,0 +1,17 @@
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        left = 0
+        right = len(height) - 1
+        
+        max_area = float("-inf")
+        while left < right:
+            if height[left] <= height[right]:
+                area = height[left] * (right - left)
+                max_area = max(max_area, area)
+                left += 1
+            else:
+                area = height[right] * (right - left)
+                max_area = max(max_area, area)
+                right -= 1
+        
+        return max_area