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

Time Complexity: Container With Most Water

O(n)

Understanding Time Complexity

We want to know how the time needed to find the container with the most water grows as the number of lines increases.

How does the number of steps change when we have more lines to check?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn max_area(height: Vec<i32>) -> i32 {
    let (mut left, mut right) = (0usize, height.len() - 1);
    let mut max_area = 0;
    while left < right {
        let width = (right - left) as i32;
        let h = if height[left] < height[right] { height[left] } else { height[right] };
        max_area = std::cmp::max(max_area, width * h);
        if height[left] < height[right] {
            left += 1;
        } else {
            right -= 1;
        }
    }
    max_area
}
    

This code finds the largest container area formed between two lines by moving pointers from both ends inward.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers inward checking areas.
• How many times: The loop runs at most once for each element, so about n times where n is the number of lines.

How Execution Grows With Input

As the number of lines grows, the steps increase roughly in a straight line.

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

Pattern observation: The steps grow directly with the number of lines, not faster or slower.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the largest container grows in a straight line with the number of lines.

Common Mistake

[X] Wrong: "We must check every pair of lines, so it takes n squared steps."

[OK] Correct: The two-pointer method smartly skips unnecessary pairs, so it only needs to check each line once.

Interview Connect

Understanding this time complexity shows you can spot efficient ways to solve problems, a skill valued in real coding challenges.

Self-Check

"What if we used a nested loop to check every pair instead of two pointers? 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 maximum area formed between two vertical lines and the x-axis

B. The minimum height in the array

C. The sum of all heights

D. The longest increasing subsequence of heights

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, which means maximizing the area between them.

2. Step 2: Identify what to calculate

   The area is calculated by the distance between the two lines multiplied by the smaller height of the two lines.

3. Final Answer:

   The maximum area formed between two vertical lines and the x-axis -> Option A

4. Quick Check:

   Container with most water = max area between lines [OK]

Hint: Look for two lines maximizing width x min height [OK]

Common Mistakes:

• Confusing sum of heights with area
• Choosing tallest lines without considering distance
• Ignoring the smaller height between two lines

2. Which Rust code snippet correctly declares a vector of heights for the Container With Most Water problem?

easy

A. let heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};

B. let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];

C. let heights = (1, 8, 6, 2, 5, 4, 8, 3, 7);

D. let heights = [1, 8, 6, 2, 5, 4, 8, 3, 7];

Solution

1. Step 1: Identify Rust vector syntax

   In Rust, a vector is declared using vec![...] macro to create a growable array.

2. Step 2: Check each option

   let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7]; uses vec![...] correctly. let heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]; is an array, not a vector. let heights = (1, 8, 6, 2, 5, 4, 8, 3, 7); is a tuple. let heights = {1, 8, 6, 2, 5, 4, 8, 3, 7}; uses braces which is invalid syntax.

3. Final Answer:

   let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7]; -> Option B

4. Quick Check:

   Vector syntax = vec![...] [OK]

Hint: Use vec![...] for vectors in Rust [OK]

Common Mistakes:

• Using array syntax instead of vector
• Using tuple parentheses instead of brackets
• Using braces which is invalid

3. What is the output of this Rust code snippet that finds the max water container area?

fn max_area(height: Vec<i32>) -> i32 {
    let mut left = 0;
    let mut right = height.len() - 1;
    let mut max_area = 0;
    while left < right {
        let width = (right - left) as i32;
        let h = if height[left] < height[right] { height[left] } else { height[right] };
        let area = width * h;
        if area > max_area {
            max_area = area;
        }
        if height[left] < height[right] {
            left += 1;
        } else {
            right -= 1;
        }
    }
    max_area
}

fn main() {
    let heights = vec![1,8,6,2,5,4,8,3,7];
    println!("{}", max_area(heights));
}

medium

A. 48

B. 56

C. 64

D. 49

Solution

1. Step 1: Understand the two-pointer approach

   Start with left at 0 and right at last index, calculate area, move pointer with smaller height inward.

2. Step 2: Trace the code with given heights

   Max area found is between height 8 at index 1 and height 7 at index 8, width = 7, area = 7 * 7 = 49.

3. Final Answer:

   49 -> Option D

4. Quick Check:

   Max area = 49 [OK]

Hint: Use two pointers, move smaller height inward [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not updating max area correctly
• Moving wrong pointer in the loop

4. Identify the bug in this Rust function for Container With Most Water and select the fix:

fn max_area(height: Vec<i32>) -> i32 {
    let mut left = 0;
    let mut right = height.len();
    let mut max_area = 0;
    while left < right {
        let width = (right - left) as i32;
        let h = if height[left] < height[right] { height[left] } else { height[right] };
        let area = width * h;
        if area > max_area {
            max_area = area;
        }
        if height[left] < height[right] {
            left += 1;
        } else {
            right -= 1;
        }
    }
    max_area
}

medium

A. Change right initialization to height.len() - 1

B. Change while condition to left <= right

C. Use max instead of min height for area calculation

D. Remove the if condition inside the loop

Solution

1. Step 1: Check initialization of right pointer

   Rust vectors are zero-indexed, so last index is length - 1, not length.

2. Step 2: Identify impact of wrong right pointer

   Using length causes out-of-bounds access when accessing height[right], leading to runtime error.

3. Final Answer:

   Change right initialization to height.len() - 1 -> Option A

4. Quick Check:

   Right pointer = last index = len() - 1 [OK]

Hint: Index last element with len() - 1, not len() [OK]

Common Mistakes:

• Using length as index causing panic
• Changing loop condition incorrectly
• Using max height instead of min for area

5. Given a vector of heights, how would you modify the two-pointer approach to also return the indices of the two lines forming the container with the most water?

hard

A. Return the sum of the two heights instead of area

B. Use a nested loop to check all pairs and store indices of max area

C. Keep track of max_area and update a tuple (left_index, right_index) whenever max_area changes

D. Only return the left pointer index as the answer

Solution

1. Step 1: Understand the two-pointer approach

   We move pointers inward and calculate area at each step, updating max area found.

2. Step 2: Track indices when max area updates

   Along with max_area, store the current left and right indices whenever a larger area is found.

3. Final Answer:

   Keep track of max_area and update a tuple (left_index, right_index) whenever max_area changes -> Option C

4. Quick Check:

   Track indices with max area updates [OK]

Hint: Update indices when max area changes [OK]

Common Mistakes:

• Returning sum instead of area
• Using nested loops losing O(n) efficiency
• Returning only one index instead of both