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Commit 643b007

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Merge pull request #363 from MoigeMatino/container-with-most-water
feat: container with most water
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## **Problem Statement**
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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])`.
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Find two lines that together with the x-axis form a container, such that the container contains the most water.
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Return the maximum amount of water a container can store.
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Notice that you may not slant the container.
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### Example 1
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Input: height = [1,8,6,2,5,4,8,3,7]
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Output: 49
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Explanation: The vertical lines are represented by array [1,8,6,2,5,4,8,3,7].
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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.
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To get the width: 8 - 1 = 7
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To get height = min(8, 7) = 7
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To get area = height * width = 7 * 7 = 49
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### Example 2
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Input: height = [1,1]
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Output: 1

‎arrays_and_strings/container_with_most_water/__init__.py

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