Container With Most Water

Brute Force

Iterate through the entire heights array and compute the area for each possibility - $O(n^2)$

• you’ll notice that if we start at opposite ends, that is the largest length we can get
• the largest height is also bottlenecked by the smaller height

Optimal 2 Pointer Approach

• Start at opposite ends
• Calculate the height
• Change the pointer of the bottlenecking height, ensure we are keeping the larger height

class Solution:
    def maxArea(self, height: List[int]) -> int:
        l, r = 0, len(height) - 1
        maxArea = 0

        while l < r:
            # compute the height
            area = min(height[l], height[r]) * (r-l)
            maxArea = max(maxArea, area)

            # inc/dec the bottleneck
            if height[l] < height[r]:
                l += 1
            else: 
                r -= 1

        return maxArea