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

Recall & Review

beginner

What is the main goal of the "Container With Most Water" problem?

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 optimal solution for "Container With Most Water"?

Two pointers start at both ends and move inward to find the maximum area efficiently by comparing heights and narrowing the search space.

beginner

What determines the area of water between two lines in the problem?

The area is determined by the shorter line's height multiplied by the distance between the two lines.

intermediate

In the two-pointer approach, which pointer do we move and why?

We move the pointer at the shorter line inward because moving the taller line won't increase the area, but moving the shorter line might find a taller line.

intermediate

What is the time complexity of the optimal two-pointer 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 height of the container depend on in the "Container With Most Water" problem?

A The taller of the two lines

B The first line in the array

C The average height of the two lines

D The shorter of the two lines

The container's height is limited by the shorter line because water cannot go above it.

What is the initial position of the two pointers in the optimal solution?

A Both at the end of the array

B One at the start and one at the end of the array

C Both at the start of the array

D Random positions in the array

One pointer starts at the beginning and the other at the end to check the widest container first.

Why do we move the pointer at the shorter line instead of the taller line?

A Because moving the shorter line might find a taller line and increase area

B Because the problem requires moving the left pointer only

C Because moving either pointer has the same effect

D Because moving the taller line always decreases area

Moving the shorter line pointer may find a taller line that can hold more water.

What is the time complexity of the brute force solution for this problem?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

The brute force checks all pairs, resulting in O(n^2) time.

Which of these best describes the two-pointer approach?

A Move pointers inward based on line heights

B Use recursion to find max area

C Check all pairs of lines

D Sort the array first

The two-pointer approach moves pointers inward, choosing which pointer to move based on the shorter line.

Explain how the two-pointer technique works to solve the "Container With Most Water" problem efficiently.

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

Describe why the area is limited by the shorter line between two lines in the container problem.

Imagine water spilling over the shorter line if it were 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 of water that can be contained between two lines.

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.

2. Step 2: Identify what to calculate

   We calculate the area between two lines as the width times the smaller height.

3. Final Answer:

   The maximum area of water that can be contained between two lines. -> Option A

4. Quick Check:

   Container area = width * min(height1, height2) [OK]

Hint: Look for two lines maximizing width and height [OK]

Common Mistakes:

• Thinking it's about sum of heights
• Confusing maximum area with minimum height
• Ignoring width between lines

2. Which of the following is the correct way to declare the main method in a C# program that solves the Container With Most Water problem?

easy

A. void main()

B. public static void Main(string[] args)

C. static Main()

D. public void Main(string args)

Solution

1. Step 1: Recall C# main method syntax

   The main method must be public static void and accept string array args.

2. Step 2: Match options with correct syntax

   public static void Main(string[] args) matches the correct signature: public static void Main(string[] args).

3. Final Answer:

   public static void Main(string[] args) -> Option B

4. Quick Check:

   Main method signature = public static void Main(string[] args) [OK]

Hint: Main method needs public static void with string[] args [OK]

Common Mistakes:

• Using void main() without public static
• Missing string array brackets
• Incorrect parameter type

3. What is the output of the following C# code snippet?

int[] height = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int maxArea = 0;
int left = 0, right = height.Length - 1;
while (left < right) {
    int area = Math.Min(height[left], height[right]) * (right - left);
    if (area > maxArea) maxArea = area;
    if (height[left] < height[right]) left++; else right--;
}
Console.WriteLine(maxArea);

medium

A. 48

B. 49

C. 56

D. 64

Solution

1. Step 1: Understand the two-pointer approach

   Start with left=0 and right=8, calculate area and move pointers inward choosing smaller height side.

2. Step 2: Trace key iterations

   Max area found is 49 between height[1]=8 and height[8]=7 with width 7.

3. Step 3: Continue tracing

   However, a larger area of 56 is found between height[1]=8 and height[6]=8 with width 7.

4. Final Answer:

   56 -> Option C

5. Quick Check:

   Max area = 56 [OK]

Hint: Use two pointers from ends, move smaller height inward [OK]

Common Mistakes:

• Calculating area incorrectly
• Moving wrong pointer
• Not updating maxArea properly

4. Identify the bug in this C# code snippet for Container With Most Water:

int maxArea = 0;
int left = 0, right = height.Length - 1;
while (left <= right) {
    int area = Math.Min(height[left], height[right]) * (right - left);
    if (area > maxArea) maxArea = area;
    if (height[left] < height[right]) left++;
    else right--;
}
return maxArea;

medium

A. maxArea should be initialized to -1 instead of 0.

B. The area calculation should use Math.Max instead of Math.Min.

C. The pointers should both move inward every iteration.

D. The loop condition should be 'left < right' instead of 'left <= right'.

Solution

1. Step 1: Check loop condition

   Using 'left <= right' causes area calculation when left == right, width zero, unnecessary.

2. Step 2: Correct loop condition

   Change to 'left < right' to avoid zero width calculation.

3. Final Answer:

   The loop condition should be 'left < right' instead of 'left <= right'. -> Option D

4. Quick Check:

   Loop must run while left < right [OK]

Hint: Loop until pointers cross, not equal [OK]

Common Mistakes:

• Using <= causes zero width area
• Using Math.Max instead of Math.Min for height
• Moving both pointers each time

5. Given an array of heights, how would you modify the two-pointer approach to find the container with the second largest water area?

hard

A. Track the top two max areas during the pointer moves and return the second largest.

B. Sort the array and pick the two largest heights.

C. Use a nested loop to check all pairs and pick the second largest area.

D. Return the area of the container formed by the first and last lines.

Solution

1. Step 1: Use two-pointer approach with tracking

   While moving pointers, keep track of the largest and second largest areas found.

2. Step 2: Update max and second max areas properly

   When a new area is larger than max, update second max to old max, max to new area; else if area is between max and second max, update second max.

3. Final Answer:

   Track the top two max areas during the pointer moves and return the second largest. -> Option A

4. Quick Check:

   Track max and second max areas during iteration [OK]

Hint: Keep two max values while scanning with two pointers [OK]

Common Mistakes:

• Sorting loses original indices and width info
• Nested loops are inefficient
• Returning first and last lines area only