Container With Most Water in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the maximum area variable initialized to zero.

DSA C++

int maxArea = [1];

Drag options to blanks, or click blank then click option'

A0

BINT_MAX

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing maxArea to 1 or a negative number.

Using INT_MAX which is too large and incorrect here.

We start with maxArea as 0 because initially we haven't calculated any area.

2fill in blank

medium

Complete the code to calculate the width between two pointers left and right.

DSA C++

int width = [1] - left;

Drag options to blanks, or click blank then click option'

AmaxArea

Bleft

Cright

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using left - right which can be negative.

Using maxArea or min which are unrelated here.

Width is the distance between the right and left pointers, so right - left.

3fill in blank

hard

Fix the error in the code to update maxArea with the maximum of current maxArea and new area.

DSA C++

maxArea = std::[1](maxArea, area);

Drag options to blanks, or click blank then click option'

Amin

Bmax

Cabs

Dswap

Attempts:

3 left

💡 Hint

Common Mistakes

Using std::min which gives the smaller value.

Using std::abs or std::swap which are unrelated.

We use std::max to keep the larger value between maxArea and area.

4fill in blank

hard

Fill both blanks to move the pointer pointing to the smaller height inward.

DSA C++

if (height[left] < height[right]) {
    left [1];
} else {
    right [2];
}

Drag options to blanks, or click blank then click option'

A++

B--

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Moving both pointers in the same direction.

Using += or -= without a value.

If left height is smaller, move left pointer forward (++). Otherwise, move right pointer backward (--).

5fill in blank

hard

Fill all three blanks to complete the function that returns the maximum water container area.

DSA C++

int maxAreaContainer(const std::vector<int>& height) {
    int left = 0, right = height.size() [1] 1;
    int maxArea = 0;
    while (left < right) {
        int width = right [2] left;
        int area = std::min(height[left], height[right]) * width;
        maxArea = std::max(maxArea, area);
        if (height[left] < height[right]) {
            left [3];
        } else {
            right--;
        }
    }
    return maxArea;
}

Drag options to blanks, or click blank then click option'

A-

C++

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Setting right to size + 1 or size.

Using + instead of - for width calculation.

Using -- instead of ++ for left pointer increment.

We set right to size - 1, calculate width as right - left, and increment left pointer when left height is smaller.

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