Container With Most Water in DSA C - Time & Space Complexity

Time Complexity: Container With Most Water

O(n)

Understanding Time Complexity

We want to find how fast the algorithm finds the container holding the most water.

How does the time taken grow as the number of lines (heights) increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the largest water container by moving two pointers from both ends towards the center.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves pointers inward.
• How many times: At most, it runs once for each element, so about n times.

How Execution Grows With Input

As the number of lines increases, the loop runs roughly once per line.

Input Size (n)	Approx. Operations
10	About 10
100	About 100
1000	About 1000

Pattern observation: The operations grow linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows directly in proportion to the number of lines.

Common Mistake

[X] Wrong: "Because there are two pointers, the time is O(n²)."

[OK] Correct: Each pointer moves only forward or backward once, so total steps are at most n, not n times n.

Interview Connect

Understanding this linear time approach shows you can optimize by avoiding unnecessary checks, a key skill in problem solving.

Self-Check

"What if we used a nested loop to check every pair? How would the time complexity change?"

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