Container with Most Water

solution.py

'''
  Explanation:
    - Intuitvely, we can find the possible areas for each pair of heights
    - For each pair, we check the area by multiplying the width to the height without spill over. We need to consider the minimum height between the pair.
    - However, can we reduce the number of times we find the possible areas for every pair of the heights which is caused by the minimum height in each pair?
      - Yes, we can. We can have two pointers at both endpoints and perform the area of the ractangle. 
      - When the height is minimum for a left pointer, we can move to the pointer to the right and when the height is minimum for a right pointer, we move the pointer to the left to the right. This is done to look for the next possible maximum height. 
    - TC: O(n), SC: O(1)
'''

class Solution:
    def maxArea(self, height: List[int]) -> int:
        ans = 0
        i = 0
        j = len(height) - 1
        
        while i < j:
            area_of_rectangle = (j - i) * (min(height[i], height[j]))
            ans = max(ans, area_of_rectangle)
            
            if height[i] < height[j]:
                i += 1
            else: j -= 1
                
        return ans