LeetCode: Container with Most Water

Problem:

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

Code:

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

        while left<right:
            w = right-left
            h = min(height[left], height[right])
            max_area = max(w*h, max_area)

            if height[left]<=height[right]:
                left+=1
            else:
                right-=1

        return max_area

Key Points:

• The container area is controlled by: width = distance between pointers height = min(height[left], height[right]) area = width * height

• Start two pointers: left = 0 # beginning of array right = n - 1 # end of array

• At each step:

  • Compute current area = (right - left) * min(height[left], height[right])

  • Update max_area if this area is larger.

• Pointer movement rule:

  • Move the pointer with the smaller height.

  • Because the height limits the area, and moving the bigger one won’t help.

  • This greedy decision ensures we explore only the useful possibilities.

• Stop when both pointers meet.

• Why two pointers work:

  • Maximum width starts at ends → best chance for big area.

  • Only way to get a bigger area is to try for a taller height, so move the smaller wall inward.

• Time Complexity: O(n) → one pass

• Space Complexity: O(1) → constant extra memory