Container With Most Water in DSA C# - Time & Space Complexity

Time Complexity: Container With Most Water

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the container with the most water changes as the number of lines increases.

How does the number of operations grow when we have more lines to check?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the largest amount of water that can be held between two lines represented by the array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers from the ends towards the center.
• How many times: The loop runs at most once for each element, so about n times where n is the number of lines.

How Execution Grows With Input

As the number of lines increases, the number of steps grows roughly the same as the number of lines.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase linearly as the input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the largest container grows in direct proportion to the number of lines.

Common Mistake

[X] Wrong: "Checking every pair of lines is necessary, so the time is always O(n²)."

[OK] Correct: The two-pointer method smartly skips unnecessary pairs, so it only needs to check each line once, making it much faster.

Interview Connect

Understanding this approach shows you can find efficient solutions by reducing unnecessary work, a skill highly valued in problem solving.

Self-Check

"What if we used a nested loop to check all pairs instead of two pointers? How would the time complexity change?"

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