Container With Most Water in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Container With Most Water Calculation

What is the output of the following code that calculates the maximum water container area?

DSA Python

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0
    while left < right:
        width = right - left
        current_height = min(height[left], height[right])
        max_water = max(max_water, width * current_height)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_water

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

A 56

B 49

C 48

D 36

💡 Hint

Think about the width and the minimum height between two lines.

The maximum area is found between the lines at index 1 and 8 with heights 8 and 7, width 7, so area = 7 * 7 = 49.

🧠 Conceptual

intermediate

Understanding the Two Pointer Approach

Why does the two-pointer approach work efficiently for the Container With Most Water problem?

A Because it checks all pairs of lines to find the maximum area.

B Because it sorts the array first to find the tallest lines.

C Because it uses recursion to explore all possible containers.

D Because it moves pointers inward, always discarding the shorter line to find a potentially taller line and larger area.

💡 Hint

Think about why moving the shorter line pointer might help.

The two-pointer method moves inward from both ends, discarding the shorter line because the area is limited by the shorter line, hoping to find a taller line that can increase area.

🔧 Debug

advanced

Identify the Error in Container Area Calculation

What error does the following code raise when executed?

DSA Python

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0
    while left < right:
        width = right - left
        current_height = max(height[left], height[right])
        max_water = max(max_water, width * current_height)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_water

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

A It raises an IndexError.

B It raises a TypeError.

C It returns the wrong maximum area value.

D It runs correctly and returns 49.

💡 Hint

Check how the height is chosen for area calculation.

The code uses max() instead of min() to find the limiting height, which causes it to overestimate the area and return a wrong value.

❓ Predict Output

advanced

Output of Container With Most Water with Equal Heights

What is the output of this code when all heights are equal?

DSA Python

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0
    while left < right:
        width = right - left
        current_height = min(height[left], height[right])
        max_water = max(max_water, width * current_height)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_water

print(max_area([5,5,5,5,5]))

A 20

B 25

C 15

D 10

💡 Hint

Calculate area using width and height for equal heights.

The maximum area is between the first and last line: width = 4, height = 5, area = 20.

🚀 Application

expert

Maximum Water Container with Large Input

Given a list of 100,000 heights all set to 1, what is the maximum water container area?

A 99999

B 0

C 1

D 100000

💡 Hint

Consider the width between the first and last line and the minimum height.

The maximum area is the width between the first and last line (99999) times the height (1), so 99999.