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+## **Problem Statement**
+
+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
+ Input: height = [1,8,6,2,5,4,8,3,7]
+ Output: 49
+
+ Explanation: The vertical lines are represented by array [1,8,6,2,5,4,8,3,7].
+ In this case, the max area of water the container can contain is 49. Derived from (1, height[1]) and (8, height[8]) which gives us the heights; h1 = 8, h2 = 7.
+ To get the width: 8 - 1 = 7
+ To get height = min(8, 7) = 7
+ To get area = height * width = 7 * 7 = 49
+
+### Example 2
+ Input: height = [1,1]
+ Output: 1
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diff --git a/arrays_and_strings/container_with_most_water/__init__.py b/arrays_and_strings/container_with_most_water/__init__.py
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diff --git a/arrays_and_strings/container_with_most_water/container_with_most_water.py b/arrays_and_strings/container_with_most_water/container_with_most_water.py
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+def max_area(height: list[int]) -> int:
+ """
+ Calculate the maximum area of water a container can store, formed by two lines
+ from the given list of heights and the x-axis.
+
+ Parameters:
+ height (list[int]): A list of integers representing the heights of vertical lines.
+
+ Returns:
+ int: The maximum area of water that can be contained.
+
+ """
+ left = 0
+ right = len(height) - 1
+ max_area = 0
+
+ while left < right:
+ # Calculate the height and width of the current container
+ curr_height = min(height[left], height[right])
+ curr_width = right - left
+ curr_area = curr_height * curr_width
+
+ # Update max_area if the current area is larger
+ max_area = max(curr_area, max_area)
+
+ # Move the pointer pointing to the shorter line
+ if height[left] < height[right]:
+ left += 1
+ else:
+ right -= 1
+
+ return max_area
+
+# Approach and Reasoning:
+# -----------------------
+# - We use the two-pointer technique to solve this problem efficiently.
+# - Initialize two pointers, `left` at the start and `right` at the end of the list.
+# - The width of the container is the distance between the two pointers.
+# - The height of the container is determined by the shorter of the two lines at the pointers.
+# - Calculate the area for the current pair of lines and update `max_area` if this area is larger.
+# - Move the pointer pointing to the shorter line inward to potentially find a taller line,
+# which might result in a larger area.
+# - Repeat the process until the two pointers meet.
+
+# Time Complexity:
+# ----------------
+# - The time complexity of this approach is O(n), where n is the number of elements in the `height` list.
+# - This is because each element is processed at most once as the pointers move towards each other.
+
+# Space Complexity:
+# -----------------
+# - The space complexity is O(1) because we are using a constant amount of extra space,
+# regardless of the input size.
+
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