Container With Most Water in DSA C# - Execution Trace

Concept Flow - Container With Most Water

Start with two pointers

↓

Calculate area between pointers

↓

Compare with max area

↓

Move pointer at shorter line inward

↓

Repeat until pointers meet

↓

Return max area found

We use two pointers at the ends of the array, calculate area, update max, then move the shorter pointer inward until they meet.

Execution Sample

DSA C#

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

Calculates the maximum water container area by moving two pointers inward.

Execution Table

Step	Operation	Left Pointer	Right Pointer	Height[left]	Height[right]	Width	Area	Max Area	Pointer Moved	Visual State
1	Initialize pointers	0	8	1	7	8	0	0	None	[1,8,6,2,5,4,8,3,7]
2	Calculate area	0	8	1	7	8	8	8	Left++	[1,8,6,2,5,4,8,3,7]
3	Calculate area	1	8	8	7	7	49	49	Right--	[1,8,6,2,5,4,8,3,7]
4	Calculate area	1	7	8	3	6	18	49	Right--	[1,8,6,2,5,4,8,3,7]
5	Calculate area	1	6	8	8	5	40	49	Right--	[1,8,6,2,5,4,8,3,7]
6	Calculate area	1	5	8	4	4	16	49	Right--	[1,8,6,2,5,4,8,3,7]
7	Calculate area	1	4	8	5	3	15	49	Right--	[1,8,6,2,5,4,8,3,7]
8	Calculate area	1	3	8	2	2	4	49	Right--	[1,8,6,2,5,4,8,3,7]
9	Calculate area	1	2	8	6	1	6	49	Right--	[1,8,6,2,5,4,8,3,7]
10	Calculate area	1	1	8	8	0	0	49	Pointers meet	[1,8,6,2,5,4,8,3,7]

💡 Pointers meet at index 1, loop ends.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	Final
left	0	1	1	1	1	1	1	1	1	1	1
right	8	8	7	6	5	4	3	2	1	1	1
max	0	8	49	49	49	49	49	49	49	49	49

Key Moments - 3 Insights

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

Because the area is limited by the shorter line, moving the taller pointer won't increase area. See steps 2 and 3 in execution_table where left pointer moves because height[left] < height[right].

Why does the loop stop when left equals right?

When pointers meet, width becomes zero, so no container can be formed. The exit_note and step 10 in execution_table show this.

How is max area updated during the process?

At each step, area is calculated and compared with max. If larger, max updates. See max column in execution_table rows 2 and 3 where max changes from 8 to 49.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the max area?

A 8

B 18

C 49

D 40

💡 Hint

Check the 'Max Area' column at step 3 in execution_table.

At which step does the right pointer first move?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Pointer Moved' column in execution_table to see when 'Right--' first appears.

If height[left] was always greater than height[right], which pointer would move?

A Right pointer moves

B Left pointer moves

C Both pointers move simultaneously

D No pointer moves

💡 Hint

Refer to the if condition in the code and pointer movement logic in execution_table.

Concept Snapshot

Container With Most Water:
- Use two pointers at array ends
- Calculate area = min(height[left], height[right]) * width
- Update max area if larger
- Move pointer at shorter height inward
- Repeat until pointers meet
- Return max area found

Full Transcript

This concept uses two pointers starting at the ends of the height array. At each step, it calculates the area formed by the lines at these pointers and the distance between them. The maximum area found is updated if the current area is larger. Then, the pointer at the shorter line moves inward to try to find a larger area. This repeats until the pointers meet, at which point the maximum area is returned.

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