Container With Most Water in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Two Pointer Approach for Container With Most Water

What is the output of the following C code that finds the maximum water container area using two pointers?

DSA C

int maxArea(int* height, int heightSize) {
    int left = 0, right = heightSize - 1;
    int max_area = 0;
    while (left < right) {
        int width = right - left;
        int min_height = height[left] < height[right] ? height[left] : height[right];
        int area = width * min_height;
        if (area > max_area) max_area = area;
        if (height[left] < height[right])
            left++;
        else
            right--;
    }
    return max_area;
}

int main() {
    int height[] = {1,8,6,2,5,4,8,3,7};
    int size = sizeof(height)/sizeof(height[0]);
    int result = maxArea(height, size);
    printf("%d\n", result);
    return 0;
}

A 56

B 48

C 50

D 49

💡 Hint

Think about the maximum area formed between two lines by moving pointers inward.

The maximum area is formed between height[1] = 8 and height[8] = 7 with width 7, so area = 7 * 7 = 49.

❓ Predict Output

intermediate

Output of Brute Force Approach for Container With Most Water

What is the output of this brute force C code that calculates the maximum water container area?

DSA C

int maxArea(int* height, int heightSize) {
    int max_area = 0;
    for (int i = 0; i < heightSize; i++) {
        for (int j = i + 1; j < heightSize; j++) {
            int width = j - i;
            int min_height = height[i] < height[j] ? height[i] : height[j];
            int area = width * min_height;
            if (area > max_area) max_area = area;
        }
    }
    return max_area;
}

int main() {
    int height[] = {1,1};
    int size = sizeof(height)/sizeof(height[0]);
    int result = maxArea(height, size);
    printf("%d\n", result);
    return 0;
}

A 1

B 0

C 2

D 3

💡 Hint

Check the area between the only two lines.

There are only two lines both of height 1 and distance 1 apart, so area = 1 * 1 = 1.

🔧 Debug

advanced

Identify the Error in Two Pointer Container Code

What error will this C code produce when compiled and run?

DSA C

int maxArea(int* height, int heightSize) {
    int left = 0, right = heightSize - 1;
    int max_area = 0;
    while (left < right) {
        int width = right - left;
        int min_height;
        if (height[left] < height[right])
            min_height = height[left];
        else
            min_height = height[right];
        int area = width * min_height;
        if (area > max_area) max_area = area;
        if (height[left] < height[right])
            left++;
        else
            right--;
    }
    return max_area;
}

int main() {
    int height[] = {1,8,6,2,5,4,8,3,7};
    int size = sizeof(height)/sizeof(height[0]);
    int result = maxArea(height, size);
    printf("%d\n", result);
    return 0;
}

A Compilation error: missing semicolon

B No error, outputs 49

C Runtime error: segmentation fault

D Logic error: returns 0

💡 Hint

Check variable initialization and syntax carefully.

The code is syntactically correct and logically implements the two pointer approach correctly. It outputs 49.

🧠 Conceptual

advanced

Why Two Pointer Approach is Efficient for Container With Most Water?

Why does the two pointer approach work efficiently to find the maximum water container area?

A Because it sorts the array first to find the largest heights

B Because it checks all pairs of lines to find the maximum area

C Because it moves pointers inward, discarding smaller heights to maximize area quickly

D Because it uses recursion to explore all combinations

💡 Hint

Think about how moving pointers affects the possible area.

The two pointer approach moves inward from the widest container, discarding the smaller height pointer to find larger areas efficiently without checking all pairs.

🚀 Application

expert

Maximum Water Container Area for Large Input

Given an array of 1,000,000 heights where all heights are 1 except the first and last which are 10, what is the maximum water container area?

A 9,999,990

B 10

C 1,000,000

D 20

💡 Hint

Consider the width and the minimum height between the first and last lines.

The maximum area is between the first and last lines: width = 999,999 and min height = 10, so area = 999,999 * 10 = 9,999,990.