Container With Most Water in DSA Rust - Build from Scratch

Container With Most Water

📖 Scenario: You are helping a water storage company to find the best container shape. They have vertical lines representing walls of different heights. You want to find two walls that together with the x-axis form a container that holds the most water.

🎯 Goal: Build a Rust program that finds the maximum amount of water that can be contained between two vertical lines in a list of heights.

📋 What You'll Learn

Create a vector called heights with the exact values [1, 8, 6, 2, 5, 4, 8, 3, 7].

Create two variables left and right to represent the start and end indices of the vector.

Use a while loop to move left and right pointers to find the maximum water container.

Print the maximum water amount found.

💡 Why This Matters

🌍 Real World

This problem models how to find the best container shape to hold the most water, useful in design and engineering.

💼 Career

Understanding two-pointer techniques and optimization is important for software engineering interviews and real-world algorithmic problem solving.

1

Create the heights vector

Create a vector called heights with the exact values [1, 8, 6, 2, 5, 4, 8, 3, 7].

DSA Rust

# Create the heights vector with the given values
// Your code here

Hint

Use let heights = vec![...] to create the vector.

2

Initialize left and right pointers

Create two variables called left and right. Set left to 0 and right to the last index of heights.

DSA Rust

fn main() {
    let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];
    // Initialize left and right pointers
    // Your code here
}

Hint

Use let mut to create variables that will change.

3

Find the maximum water container

Create a variable called max_area and set it to 0. Use a while loop with the condition left < right. Inside the loop, calculate the current area using the minimum height between heights[left] and heights[right] multiplied by the distance between right and left. Update max_area if the current area is larger. Move left forward if heights[left] is less than heights[right], otherwise move right backward.

DSA Rust

fn main() {
    let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];
    let mut left = 0;
    let mut right = heights.len() - 1;
    // Calculate max_area using two pointers and a while loop
    // Your code here
}

Hint

Use while left < right and update max_area inside the loop.

4

Print the maximum water area

Add a println! statement to display the value of max_area.

DSA Rust

fn main() {
    let heights = vec![1, 8, 6, 2, 5, 4, 8, 3, 7];
    let mut left = 0;
    let mut right = heights.len() - 1;
    let mut max_area = 0;

    while left  max_area {
            max_area = current_area;
        }
        if heights[left] < heights[right] {
            left += 1;
        } else {
            right -= 1;
        }
    }
    // Print the max_area
    // Your code here
}

Hint

Use println!("{}", max_area); 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 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