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.

Practice

1. What does the problem Container With Most Water ask you to find given an array of heights?

easy

A. The maximum area formed between two vertical lines and the x-axis

B. The minimum height in the array

C. The sum of all heights in the array

D. The number of lines taller than a given height

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two lines that together with the x-axis form a container holding the most water, which means the maximum area between two vertical lines.

2. Step 2: Identify what to calculate

   We calculate the area as the width between two lines multiplied by the smaller height of the two lines.

3. Final Answer:

   The maximum area formed between two vertical lines and the x-axis -> Option A

4. Quick Check:

   Container with most water = maximum area [OK]

Hint: Look for max area between two heights, not sum or count [OK]

Common Mistakes:

• Confusing area with sum of heights
• Looking for minimum height only
• Counting lines instead of calculating area

2. Which of the following is the correct way to initialize two pointers for the Container With Most Water problem in Python?

easy

A. left, right = 1, len(height)

B. left, right = 0, len(height) - 1

C. left = 0; right = len(height)

D. left, right = len(height), 0

Solution

1. Step 1: Understand pointer initialization

   We start with two pointers at the edges of the array: left at 0 and right at the last index, which is length minus one.

2. Step 2: Check each option

   left, right = 0, len(height) - 1 correctly sets left to 0 and right to len(height) - 1. Others either start at wrong indices or exceed array bounds.

3. Final Answer:

   left, right = 0, len(height) - 1 -> Option B

4. Quick Check:

   Start pointers at array ends = left, right = 0, len(height) - 1 [OK]

Hint: Start pointers at first and last index [OK]

Common Mistakes:

• Setting right pointer to len(height) (out of range)
• Starting left pointer at 1 instead of 0
• Swapping left and right initialization

3. What is the output of the following Python code for the Container With Most Water problem?

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

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

medium

A. 49

B. 56

C. 64

D. 72

Solution

1. Step 1: Trace the two-pointer approach

   Start with left=0 (height=1), right=8 (height=7), area = min(1,7)*8=8. Move left pointer since height[left]

2. Step 2: Continue moving pointers and update max_area

   Next left=1 (8), right=8 (7), area = min(8,7)*7=49 (max_area updated). Then right moves left, area decreases. Max area remains 49.

3. Final Answer:

   49 -> Option A

4. Quick Check:

   Max area calculated = 49 [OK]

Hint: Move pointer with smaller height to find max area [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not updating max_area correctly
• Moving wrong pointer in the loop

4. Find the error in the following code snippet for Container With Most Water and select the fix:

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

medium

A. Increment both left and right pointers inside the loop

B. Initialize max_area to 1 instead of 0

C. Use max(height[left], height[right]) instead of min

D. Change while condition to 'left < right'

Solution

1. Step 1: Analyze the loop condition

   The loop uses 'while left <= right', which allows left and right to be equal, causing width to be zero and area zero, which is unnecessary.

2. Step 2: Correct the loop condition

   Changing to 'while left < right' ensures pointers do not cross or become equal, avoiding zero width and redundant calculations.

3. Final Answer:

   Change while condition to 'left < right' -> Option D

4. Quick Check:

   Loop until pointers cross, not equal [OK]

Hint: Use 'left < right' to avoid zero width area [OK]

Common Mistakes:

• Using <= causes zero width calculation
• Using max instead of min for height
• Moving both pointers at once

5. Given the heights array [4,3,2,1,4], what is the maximum water container area and which pair of lines form it?

hard

A. 12, formed by lines at indices 1 and 4

B. 9, formed by lines at indices 0 and 3

C. 16, formed by lines at indices 0 and 4

D. 8, formed by lines at indices 2 and 4

Solution

1. Step 1: Calculate area between lines at indices 0 and 4

   Height = min(4,4) = 4, width = 4 - 0 = 4, area = 4 * 4 = 16.

2. Step 2: Check other pairs for larger area

   Pairs (1,4): min(3,4)*3=9; (0,3): min(4,1)*3=3; (2,4): min(2,4)*2=4. None exceed 16.

3. Final Answer:

   16, formed by lines at indices 0 and 4 -> Option C

4. Quick Check:

   Max area = 16 from edges [OK]

Hint: Check edges first, then move inward [OK]

Common Mistakes:

• Ignoring width when calculating area
• Using max height instead of min
• Not checking all pairs properly