Code-Memo

Container With Most Water

Given ( n ) non-negative integers ( a_1, a_2, \ldots, a_n ), where each represents a point at coordinate ((i, a_i)). ( n ) vertical lines are drawn such that the two endpoints of the line ( i ) are ((i, 0)) and ((i, a_i)). Find two lines, which together with the x-axis forms a container, such that the container contains the most water.

Example: ![[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].

Approach

Two-Pointer Technique

1. Initialization:
   • Start with two pointers, one at the beginning (left) and one at the end (right) of the array.
   • Initialize a variable area to keep track of the maximum area found.
2. Iterate and Adjust Pointers:
   • Calculate the area formed by the lines at the left and right pointers.
   • Update area with the maximum of the current area and the newly calculated area.
   • Move the pointer that points to the shorter line towards the center to potentially find a container with a larger area.
3. End Condition:
   • The loop continues until the two pointers meet.

Time Complexity

• The time complexity of this approach is ( O(n) ), where ( n ) is the length of the array. This is the optimal solution for this problem.

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        area = 0
        left, right = 0, len(height) - 1

        while left < right:
            area = max(area, min(height[left], height[right]) * (right - left))
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1

        return area