Container With Most Water

1. Clarify the problem:

   • The problem requires us to find the maximum area of water that can be trapped between two vertical lines.
   • We are given an array of non-negative integers representing the heights of the vertical lines.
   • The width of each vertical line is considered to be 1 unit.
   • We need to determine the maximum area that can be formed by selecting any two vertical lines from the given array.

2. Analyze the problem:

   • Input: An array of non-negative integers representing the heights of vertical lines.
   • Output: The maximum area of water that can be trapped.
   • Constraints:
     • The input array can contain up to 10^5 elements.
     • The height of each vertical line can range from 0 to 10^4.

3. Design an algorithm:

   • We can solve this problem using a two-pointer approach.
   • We start with two pointers, one at the beginning of the array (left pointer) and one at the end (right pointer).
   • The width between the two pointers represents the maximum width for the container.
   • We calculate the area between the two pointers using the minimum height of the two vertical lines multiplied by the width.
   • We update the maximum area if the calculated area is greater than the current maximum.
   • To maximize the area, we move the pointer with the smaller height towards the center.
   • We repeat this process until the two pointers meet.
   • Finally, we return the maximum area.

4. Explain your approach:

   • We will use a two-pointer approach to find the maximum area of water.
   • We start with two pointers, left at index 0 and right at the last index of the array.
   • We initialize a variable max_area to store the maximum area found so far.
   • While left is less than right:
     • Calculate the area between the vertical lines at left and right.
     • Update max_area if the calculated area is greater than the current max_area.
     • Move the pointer with the smaller height towards the center.
   • Finally, return the max_area.

5. Write clean and readable code:

python

def maxArea(height):
    max_area = 0
    left = 0
    right = len(height) - 1

    while left < right:
        area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, area)

        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_area

6. Test your code:

python

print(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]))  # Output: 49
print(maxArea([1, 1]))  # Output: 1
print(maxArea([4, 3, 2, 1, 4]))  # Output: 16
print(maxArea([1, 2, 1]))  # Output: 2

7. Optimize if necessary:

   • The two-pointer approach used has a time complexity of O(N), where N is the length of the input array.
   • The space complexity is O(1) since we only use a constant amount of additional space.

8. Handle error cases:

   • The code assumes the input will be a non-empty array of non-negative integers, as stated in the problem description.
   • If an empty array is passed as input, the code will return 0, indicating that there is no water trapped.

9. Discuss complexity analysis:

   • Time complexity: O(N), where N is the length of the input array.
   • Space complexity: O(1) since we only use a constant amount of additional space.