Container With Most Water in DSA C - Build from Scratch

Container With Most Water

📖 Scenario: You are helping a water storage company find the best two vertical lines to hold the most water between them. Each line's height is given, and you want to find the maximum water container that can be formed between any two lines.

🎯 Goal: Build a program that finds the maximum amount of water that can be contained between two lines represented by an array of heights.

📋 What You'll Learn

Create an array called height with the exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

Create two integer variables left and right to point to the start and end of the array

Use a while loop to move the pointers and calculate the maximum water container

Print the maximum water container value using printf

💡 Why This Matters

🌍 Real World

This problem models how to find the best container shape to hold the most water between vertical lines, useful in storage and design.

💼 Career

Understanding two-pointer techniques and array traversal is important for coding interviews and real-world algorithm optimization.

1

Create the height array

Create an integer array called height with these exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

DSA C

# Your code here

Hint

Use curly braces to list the values inside the array.

2

Initialize pointers

Create two integer variables called left and right. Set left to 0 and right to 8 (the last index of the array).

DSA C

int height[] = {1, 8, 6, 2, 5, 4, 8, 3, 7};
# Your code here

Hint

Remember the array has 9 elements, so the last index is 8.

3

Calculate maximum water container

Create an integer variable maxArea and set it to 0. Use a while loop with the condition left < right. Inside the loop, calculate the area between height[left] and height[right]. Update maxArea if the current area is larger. Move left forward if height[left] is less than height[right], otherwise move right backward.

DSA C

int height[] = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0;
int right = 8;
# Your code here

Hint

Use a loop to check all pairs by moving pointers inward.

4

Print the maximum water container

Use printf to print the value of maxArea followed by a newline.

DSA C

Hint

Use printf("%d\n", maxArea); to print the result.

Practice

1. What does the problem Container With Most Water ask you to find given an array of heights?

easy

A. The number of lines taller than a given height

B. The minimum height in the array

C. The sum of all heights in the array

D. The maximum area formed between two lines representing container walls

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 maximize

   We want to maximize the area between two lines, which is the width between them times the smaller height.

3. Final Answer:

   The maximum area formed between two lines representing container walls -> Option D

4. Quick Check:

   Max area = max water container [OK]

Hint: Focus on max area between two heights, not sum or count [OK]

Common Mistakes:

• Confusing max area with sum of heights
• Looking for minimum height only
• Counting lines instead of calculating area

2. Which of the following is the correct way to declare the two-pointer variables for solving Container With Most Water in C?

easy

A. int left = 1, right = n;

B. int left = n, right = 0;

C. int left = 0, right = n - 1;

D. int left = -1, right = n + 1;

Solution

1. Step 1: Understand pointer initialization

   We start with left pointer at the start (index 0) and right pointer at the end (index n-1) of the array.

2. Step 2: Check each option

   int left = 0, right = n - 1; correctly initializes left=0 and right=n-1. Others use invalid or out-of-bound indices.

3. Final Answer:

   int left = 0, right = n - 1; -> Option C

4. Quick Check:

   Two pointers at array ends = int left = 0, right = n - 1; [OK]

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

Common Mistakes:

• Starting pointers out of array bounds
• Using 1-based indexing in C arrays
• Swapping left and right initialization

3. What is the output of the following C code snippet for the array height = {1, 8, 6, 2, 5, 4, 8, 3, 7}?

int maxArea(int* height, int heightSize) {
    int left = 0, right = heightSize - 1, max_area = 0;
    while (left < right) {
        int area = (right - left) * (height[left] < height[right] ? height[left] : height[right]);
        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]);
    printf("%d", maxArea(height, size));
    return 0;
}

medium

A. 72

B. 49

C. 64

D. 56

Solution

1. Step 1: Initialize pointers and max_area

   left=0, right=8, max_area=0. Calculate area each step and update max_area.

2. Step 2: Trace iterations

   left=0 (1), right=8 (7), area=8*1=8, max=8, left++; left=1 (8), right=8 (7), area=7*7=49, max=49; next areas smaller (e.g. 6*3=18, 5*8=40), max_area remains 49.

3. Final Answer:

   49 -> Option B

4. Quick Check:

   Max water container area = 49 [OK]

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

Common Mistakes:

• Calculating area incorrectly as sum instead of width*height
• Not updating max_area properly
• Moving wrong pointer in the loop

4. Identify the bug in this C code snippet for Container With Most Water and select the fix:

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

medium

A. Change while condition to left < right

B. Initialize max_area to -1

C. Use max_area += area instead of max_area = area

D. Swap left and right pointers initialization

Solution

1. Step 1: Analyze loop condition

   The loop uses while(left <= right), which allows left == right, causing zero width and unnecessary calculation.

2. Step 2: Correct loop condition

   Change condition to while(left < right) to avoid zero width and infinite loop risk.

3. Final Answer:

   Change while condition to left < right -> Option A

4. Quick Check:

   Loop must run while left < right [OK]

Hint: Use strict less than for pointers to avoid zero width [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Incorrectly updating max_area by adding
• Changing pointer initialization unnecessarily

5. Given an array of heights representing vertical lines, how would you modify the two-pointer approach to find the container with the most water if the container must include the line at index 0?

hard

A. Fix left pointer at 0 and move right pointer from end towards left

B. Fix right pointer at 0 and move left pointer from start towards right

C. Use a nested loop to check all pairs including index 0

D. Use binary search to find max area including index 0

Solution

1. Step 1: Understand the constraint

   The container must include the line at index 0, so left pointer stays fixed at 0.

2. Step 2: Move right pointer from end to left

   Move right pointer from the last index towards 1, calculating area with fixed left pointer to find max area.

3. Final Answer:

   Fix left pointer at 0 and move right pointer from end towards left -> Option A

4. Quick Check:

   Fix one pointer at 0, move other to find max area [OK]

Hint: Fix one pointer, move the other to maximize area [OK]

Common Mistakes:

• Moving left pointer instead of fixing it
• Using nested loops unnecessarily
• Applying binary search incorrectly