0011.Container w Most Water

• Link: 11. Container With Most Water - Medium
• Track: Amazon

Question

Restate

• Given an integer array height, where height[i] is the height of a vertical line at i
• Choose two lines with x-axis, form a container
• Find two lines that contain the most water
• Return the maximum area

Edge Case

• len(height) < 2
• All equal heights (e.g., [5,5,5]) → widest pair is optimal

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Method	Approach	Time Complexity	Space Complexity
Method 1 ⭐	Two Pointers	$O(n)$	$O(1)$

Method 1 - Two Pointers

Approach

Key idea:

• The water area is limited by the shorter line
• To possibly increase the area, always move the pointer at the shorter line

• Use two pointers left (start) and right (end)
• At each step:
  1. Compute area using the current pair.
  2. Update res.
  3. Move the pointer at the shorter height inward:
     • Because width decreases no matter what; the only chance to get a bigger area is to potentially increase the limiting height min(...).
     • Moving the taller side cannot increase min(...) when the shorter side stays the same, so it can’t lead to a better area than the current width.

Complexity

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

Code

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

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Common Mistake

• Using while left <= right (when left == right, width is 0, no need to compute)