Container With Most Water -

https://leetcode.com/problems/container-with-most-water

A two-pointer solution is intuitive, but the question is how to not lose any candidates.

Start with two pointers on either side, and advance the pointer with the lesser height.

Algorithm

class Solution:
    # Time: O(n)
    # Space: O(1)
    def maxArea(self, height: List[int]) -> int:
        max_area = float("-inf")

        # Start with two pointers on either side
        left = 0
        right = len(height) -1

        while left < right:
            # Keep track of the max area seen
            max_area = max(max_area, (right - left) * min(height[left], height[right]))

            # Advance the pointer with the lesser height
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
        
        return max_area