Container With Most Water

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Problem

• We’re given an input array containing height values where heights[i] represent the height of the $i^{th}$ bar.

• Our function can choose any two bars to form a container. Return the maximum amount of water a container can store.

  

Example

Input: height = [1,7,2,5,4,7,3,6]
 
Output: 36

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Approach 1

Visualize the problem

Solve without code

• Look at left and right of input
• Get the min height of both since we need even heights
• Calculate the area
• Ask ourselves if the new area is great than the current
• If our left is less than our right bar then we move it by one otherwise we move right by one

Solve with code

• Create l and r pointers.
• Create a result variable to hold our current max result
• Create a while loop with the condition of while l < r
  • Calculate area by finding the width r - l and the height min(heights[r],heights[l]
  • This will give us the min height which we can multiply by the width
  • If our left pointer is smaller than our right we increase its index, otherwise subtract the right index by one
• return our result

Implementation

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

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$