Container With Most Water in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer at the start of the list.

DSA Python

left = [1]

Drag options to blanks, or click blank then click option'

A1

Blen(height)

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0.

Using length of list as starting index.

The left pointer should start at index 0, the beginning of the list.

2fill in blank

medium

Complete the code to initialize the right pointer at the end of the list.

DSA Python

right = [1]

Drag options to blanks, or click blank then click option'

A-1

B0

Clen(height)

Dlen(height) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(height) as index which is out of range.

Starting right pointer at 0.

The right pointer should start at the last index, which is length minus one.

3fill in blank

hard

Fix the error in the while loop condition to continue until left is less than right.

DSA Python

while [1] < right:

Drag options to blanks, or click blank then click option'

Aright - 1

Bleft

Cleft + 1

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right < right' which is always false.

Using 'left + 1 < right' which skips valid comparisons.

The loop should run while left pointer is less than right pointer.

4fill in blank

hard

Fill both blanks to calculate the current area and update max_area if needed.

DSA Python

current_area = (right - left) * min(height[left], height[[1]])
max_area = max(max_area, current_area) if current_area [2] max_area else max_area

Drag options to blanks, or click blank then click option'

Aleft

Bright

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using height at left pointer without min.

Updating max_area when current_area is smaller.

The height used is the minimum of heights at left and right pointers. Update max_area if current_area is greater.

5fill in blank

hard

Fill all three blanks to move the pointer with smaller height inward.

DSA Python

if height[left] [1] height[right]:
    left [2]= 1
else:
    right [3]= 1

Drag options to blanks, or click blank then click option'

A<

B+

C-

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Moving the wrong pointer based on height comparison.

Using wrong operators for pointer movement.

If height at left is less, move left pointer right by adding 1. Otherwise, move right pointer left by subtracting 1.

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