Container With Most Water in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'Container With Most Water' problem ask you to find?

It asks to find two lines in an array that, together with the x-axis, form a container that holds the maximum amount of water.

intermediate

Why do we use two pointers in the 'Container With Most Water' problem?

Two pointers start at both ends and move inward to check all possible containers efficiently, reducing time from O(n²) to O(n).

beginner

How do you calculate the area of water between two lines at positions i and j?

Area = min(height[i], height[j]) * (j - i), where height[i] and height[j] are the heights of the lines.

intermediate

What is the rule for moving pointers in the 'Container With Most Water' algorithm?

Move the pointer at the shorter line inward, because moving the taller line won't increase the area.

beginner

What is the time complexity of the optimal solution for 'Container With Most Water'?

O(n), where n is the number of lines, because each pointer moves at most n times.

What does the 'Container With Most Water' problem primarily involve?

A Finding two lines that hold the maximum water

B Sorting an array of heights

C Finding the shortest line in the array

D Calculating the sum of all heights

The problem is about finding two lines that together hold the most water.

In the two-pointer approach, where do the pointers start?

A Both at the end of the array

B One at the start and one at the end

C Both at the start of the array

D Random positions in the array

Pointers start at the two ends to check the widest container first.

How do you decide which pointer to move inward?

A Move the pointer at the taller line

B Move the pointer at the middle

C Move both pointers simultaneously

D Move the pointer at the shorter line

Moving the shorter line pointer may find a taller line and increase area.

What is the formula to calculate the water area between two lines?

A max(height[i], height[j]) * (j - i)

B height[i] + height[j]

C min(height[i], height[j]) * (j - i)

D (height[i] + height[j]) / 2

Area depends on the shorter line and the distance between lines.

What is the time complexity of the two-pointer solution?

A O(n)

B O(n log n)

C O(n²)

D O(log n)

Two pointers move inward once each, making it O(n).

Explain how the two-pointer technique works in the 'Container With Most Water' problem.

Think about how moving the shorter line pointer can help find a bigger container.

Describe how to calculate the area between two lines and why the shorter line limits the water container.

Imagine water spilling over the shorter line if it was taller.

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