Container With Most Water in DSA C# - Build from Scratch

Container With Most Water

📖 Scenario: You are helping a water storage company to find the best place to build a container between vertical lines on a flat surface. Each line's height is given, and you want to find two lines that together with the x-axis form a container that holds the most water.

🎯 Goal: Build a program that finds the maximum amount of water a container can store using the heights of vertical lines.

📋 What You'll Learn

Create an array of integers called heights with the exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

Create two integer variables called left and right to represent the start and end indices of the array

Use a while loop to move left and right pointers and calculate the maximum water container size

Print the maximum water container size as an integer

💡 Why This Matters

🌍 Real World

This problem models how to find the best container shape to hold the most water between vertical lines, useful in storage and design.

💼 Career

Understanding two-pointer techniques and array manipulation is important for coding interviews and real-world software development.

1

Create the heights array

Create an integer array called heights with these exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

DSA C#

// Create the heights array with the given values
// Your code here

Hint

Use int[] heights = { ... }; to create the array with the exact numbers.

2

Initialize pointers for the container edges

Create two integer variables called left and right. Set left to 0 and right to the last index of the heights array.

DSA C#

int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
// Initialize left and right pointers
// Your code here

Hint

Use int left = 0; and int right = heights.Length - 1; to set the pointers.

3

Calculate the maximum water container size

Create an integer variable called maxArea and set it to 0. Use a while loop that runs while left is less than right. Inside the loop, calculate the area between left and right using the minimum height of the two lines multiplied by the distance between them. Update maxArea if the current area is larger. Move the pointer that points to the shorter line inward by one.

DSA C#

int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0;
int right = heights.Length - 1;
// Calculate maxArea using a while loop
// Your code here

Hint

Use a while loop and update maxArea by comparing current area. Move the pointer of the smaller height.

4

Print the maximum water container size

Print the value of maxArea using Console.WriteLine.

DSA C#

int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0;
int right = heights.Length - 1;
int maxArea = 0;

while (left  maxArea)
    {
        maxArea = currentArea;
    }
    if (heights[left] < heights[right])
    {
        left++;
    }
    else
    {
        right--;
    }
}
// Print the maxArea
// Your code here

Hint

Use Console.WriteLine(maxArea); to print the result.

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