Container With Most Water

Problem Statement:

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return the maximum amount of water a container can store.
Notice that you may not slant the container.

Example:

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]. In this case, the max area of water (blue section) the container can contain is 49.

Intuition:

Core logic is to find the area with within two heights, i.e., area = min(heights[l], heights[r]) * (r-l)
To optimally find two heights. use two pointer approach - l=0 and r=n-1, compute area update if greater than last computed area. and for pointer updation. Check if heights[l] < heights[r] then move l else move r.
Reason is - Move the pointer pointing to the smaller height inward (because moving the larger one won’t increase area).

Code:

My Solution

def maxArea(self, height: List[int]) -> int:
	l, r = 0, len(height) - 1
	ans = 0
	while l < r:
		area = min(height[l], height[r]) * (r-l)
		ans = max(area, ans)
		if height[l] < height[r]:
			l += 1
		else:
			r -= 1
	return ans

#two-pointer