Container With Most Water in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'Container With Most Water' problem ask you to find?

It asks to find two lines in an array that, together with the x-axis, form a container that holds the maximum amount of water.

beginner

What is the main idea behind the two-pointer approach for this problem?

Use two pointers at the start and end of the array, calculate area, then move the pointer with the smaller height inward to try to find a bigger container.

intermediate

Why do we move the pointer with the smaller height in the two-pointer method?

Because the area is limited by the smaller height, moving it might find a taller line and increase the area.

intermediate

What is the time complexity of the optimal solution for 'Container With Most Water'?

O(n), where n is the number of lines, because each pointer moves at most n times.

intermediate

What would be the output for the input heights [1,8,6,2,5,4,8,3,7]?

The maximum area is 49, formed between lines at index 1 (height 8) and index 8 (height 7).

In the two-pointer approach, where do the pointers start?

A In the middle of the array

B Both at the beginning

C At the beginning and end of the array

D Both at the end

The two pointers start at the beginning and end to check the widest container first.

What limits the amount of water a container can hold?

A The shorter line

B The taller line

C The distance between lines only

D The sum of both lines' heights

The container's height is limited by the shorter line because water cannot go above it.

What happens if you move the pointer pointing to the taller line inward?

A Area might increase

B Area might decrease or stay the same

C Area always increases

D Area always decreases

Moving the taller line pointer inward might reduce width without increasing height, so area might decrease or stay the same.

What is the time complexity of the brute force solution?

A O(n^2)

B O(n)

C O(n log n)

D O(log n)

Brute force checks all pairs, which takes O(n^2) time.

Which of these is a correct step in the two-pointer algorithm?

A Move pointers randomly

B Always move the pointer with the taller line

C Move both pointers inward simultaneously

D Always move the pointer with the shorter line

Moving the pointer with the shorter line can potentially find a taller line and increase area.

Explain the two-pointer approach to solve the 'Container With Most Water' problem.

Think about how to maximize area by adjusting pointers.

Describe why the area is limited by the shorter line in the container problem.

Imagine water filling between two walls of different heights.

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