Container With Most Water in DSA C++ - Execution Trace

Concept Flow - Container With Most Water

Start with two pointers

↓

Calculate area between left and right

↓

Update max area if current area is larger

↓

Move pointer with smaller height inward

↓

left++ if height[left

↓

Repeat until left meets right

↓

Return max area found

Start with two pointers at ends, calculate area, move the smaller height pointer inward, update max area, repeat until pointers meet.

Execution Sample

DSA C++

int maxArea(vector<int>& height) {
  int left = 0, right = height.size() - 1;
  int max_area = 0;
  while (left < right) {
    int area = min(height[left], height[right]) * (right - left);
    max_area = max(max_area, area);
    if (height[left] < height[right]) left++; else right--;
  }
  return max_area;
}

Calculates the maximum water container area by moving two pointers inward based on height comparison.

Execution Table

Step	Operation	Left Pointer	Right Pointer	Height[left]	Height[right]	Area Calculated	Max Area	Pointer Moved	Visual State
1	Initialize pointers	0	8	1	3	min(1,3)*8=8	8	Move left pointer (height[left]<height[right])	[1,8,6,2,5,4,8,7,3]
2	Calculate area	1	8	8	3	min(8,3)*7=21	21	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
3	Calculate area	1	7	8	7	min(8,7)*6=42	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
4	Calculate area	1	6	8	8	min(8,8)*5=40	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
5	Calculate area	1	5	8	4	min(8,4)*4=16	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
6	Calculate area	1	4	8	5	min(8,5)*3=15	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
7	Calculate area	1	3	8	2	min(8,2)*2=4	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
8	Calculate area	1	2	8	6	min(8,6)*1=6	42	Move right pointer (height[right]<=height[left])	[1,8,6,2,5,4,8,7,3]
9	Pointers meet	1	1	8	8	Stop	42	Stop	[1,8,6,2,5,4,8,7,3]

💡 Left pointer meets right pointer, loop ends.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	Final
left	0	1	1	1	1	1	1	1	1	1
right	8	8	7	6	5	4	3	2	1	1
max_area	0	8	21	42	42	42	42	42	42	42

Key Moments - 3 Insights

Why do we move the pointer with the smaller height?

Because the area depends on the smaller height, moving the taller pointer won't increase area. See steps 1 and 2 in execution_table where moving the smaller height pointer leads to larger area.

Why does the loop stop when left meets right?

When left equals right, there is no width between pointers, so area is zero. The exit_note and step 9 in execution_table show this stopping condition.

Why do we calculate area using min(height[left], height[right]) * (right - left)?

The container's water is limited by the shorter line and the distance between pointers. This formula is used in every step's 'Area Calculated' column in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the max_area after step 3?

A 21

B 42

C 40

D 8

💡 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 1

D Step 4

💡 Hint

Look at the 'Pointer Moved' column in execution_table for the first right pointer move.

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

A Both pointers move simultaneously

B Left pointer moves

C Right pointer moves

D No pointer moves

💡 Hint

Refer to the 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]) * distance
- Move pointer with smaller height inward
- Update max area each step
- Stop when pointers meet
- Returns max water container area

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 is updated if the current area is larger. Then, the pointer at the smaller height moves inward to try to find a taller line that could increase the area. This repeats until the pointers meet, at which point the maximum area found is returned.

Practice

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

easy

A. Find two lines that together with the x-axis form a container holding the most water

B. Sort the array of heights in ascending order

C. Find the maximum height in the array

D. Calculate the sum of all heights

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two vertical lines which, together with the x-axis, form a container that holds the maximum amount of water.

2. Step 2: Identify what is being maximized

   The area between two lines is calculated by the minimum height of the two lines multiplied by the distance between them. We want to maximize this area.

3. Final Answer:

   Find two lines that together with the x-axis form a container holding the most water -> Option A

4. Quick Check:

   Goal = Max water container [OK]

Hint: Max area = min height * width between lines [OK]

Common Mistakes:

• Thinking it's about sorting heights
• Trying to sum heights instead of area
• Looking for max single height only

2. Which of the following C++ code snippets correctly initializes two pointers for the two-pointer approach in the Container With Most Water problem?

easy

A. int left = 1, right = heights.size();

B. int left = 0, right = heights.size() - 1;

C. int left = 0, right = heights.size();

D. int left = 1, right = heights.size() - 1;

Solution

1. Step 1: Understand pointer initialization

   We start with two pointers: one at the beginning (index 0) and one at the end (index size-1) of the array.

2. Step 2: Check each option

   int left = 0, right = heights.size() - 1; correctly sets left = 0 and right = size - 1, which are valid indices. Other options use invalid indices or off-by-one errors.

3. Final Answer:

   int left = 0, right = heights.size() - 1; -> Option B

4. Quick Check:

   Pointers start at ends = int left = 0, right = heights.size() - 1; [OK]

Hint: Start pointers at array ends: 0 and size-1 [OK]

Common Mistakes:

• Using size() as index (out of bounds)
• Starting pointers at 1 instead of 0
• Swapping left and right initialization

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

std::vector<int> heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0, right = heights.size() - 1;
int maxArea = 0;
while (left < right) {
    int height = std::min(heights[left], heights[right]);
    int width = right - left;
    maxArea = std::max(maxArea, height * width);
    if (heights[left] < heights[right]) {
        left++;
    } else {
        right--;
    }
}
std::cout << maxArea << std::endl;

medium

A. 56

B. 50

C. 48

D. 49

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 updating maxArea

   Next left=1 (8), right=8 (7), area= min(8,7)*7=49 (maxArea=49). Move right pointer since height[left]>height[right]. Then left=1, right=7 (3), area= min(8,3)*6=18 (maxArea=49). Move right pointer. Continue similarly until pointers meet. The maximum area found is 49.

3. Final Answer:

   49 -> Option D

4. Quick Check:

   Max area = 49 [OK]

Hint: Move pointer at smaller height to find max area [OK]

Common Mistakes:

• Not updating maxArea correctly
• Moving wrong pointer after comparison
• Calculating area with max height instead of min

4. Identify the bug in the following code snippet for the Container With Most Water problem:

int maxArea(std::vector<int>& heights) {
    int left = 0, right = heights.size();
    int max_area = 0;
    while (left < right) {
        int height = std::min(heights[left], heights[right]);
        int width = right - left;
        max_area = std::max(max_area, height * width);
        if (heights[left] < heights[right]) {
            left++;
        } else {
            right--;
        }
    }
    return max_area;
}

medium

A. The loop condition should be left <= right

B. Use max instead of min to calculate height

C. right should be initialized to heights.size() - 1

D. Increment right instead of decrementing

Solution

1. Step 1: Check pointer initialization

   right is initialized to heights.size(), which is out of bounds since valid indices go up to size()-1.

2. Step 2: Correct initialization and loop condition

   right should be heights.size() - 1 to point to the last valid element. The loop condition left < right is correct.

3. Final Answer:

   right should be initialized to heights.size() - 1 -> Option C

4. Quick Check:

   Index out of bounds fixed by right should be initialized to heights.size() - 1 [OK]

Hint: Array indices go from 0 to size-1 [OK]

Common Mistakes:

• Using size() as index causing runtime error
• Changing loop condition incorrectly
• Using max instead of min for height

5. Given an array of heights representing vertical lines, how would you modify the two-pointer approach to find the maximum water container if the array can contain zero heights (lines with zero height)?

hard

A. Skip zero height lines by moving pointers inward until non-zero heights are found

B. Treat zero heights as normal and proceed with the standard two-pointer approach

C. Remove zero height lines from the array before processing

D. Use a nested loop to check all pairs instead of two-pointer

Solution

1. Step 1: Understand zero height impact

   Lines with zero height cannot hold water, so including them wastes time and area calculations.

2. Step 2: Modify two-pointer approach

   Skip zero height lines by moving pointers inward until both point to non-zero heights. Then apply the standard two-pointer logic to maximize area.

3. Final Answer:

   Skip zero height lines by moving pointers inward until non-zero heights are found -> Option A

4. Quick Check:

   Skip zeros for max area [OK]

Hint: Ignore zero heights by moving pointers inward [OK]

Common Mistakes:

• Processing zero height lines as normal
• Removing elements changes indices and complicates logic
• Switching to nested loops loses efficiency