11. Container With Most Water

Container With Most Water - LeetCode

This medium problem becomes much simpler with a two-pointer approach.

Intuition

The area of water held by two lines depends on:

• the width between them
• the shorter of the two heights

At first glance, brute force seems natural because there are many possible pairs. But checking every pair would be too slow.

The important insight is that the shorter wall is always the limiting factor.

So if we have two pointers at the ends:

• we calculate the area they currently form
• then we move the pointer with the smaller height inward

Why? Because keeping the shorter wall and moving the taller one cannot increase the height limit. The only hope of improving the area is to find a taller shorter wall.

Implementation

I start with:

• left = 0
• right = len(height) - 1

For each step:

1. calculate the width as right - left
2. calculate the effective height as min(height[left], height[right])
3. update the best area

Then I move whichever side has the smaller height inward.

This continues until the two pointers meet. The result is the maximum area found during that process.

# 11. Container With Most Water

def maxArea(height):
    left = 0
    right = len(height) - 1
    best = 0

    while left < right:
        width = right - left
        curr_height = min(height[left], height[right])
        best = max(best, width * curr_height)

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

    return best


print(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]))