Why Container With Most Water in DSA Rust?

The Big Idea

What if you could find the biggest water container without checking every pair of walls?

The Scenario

Imagine you have a row of vertical walls of different heights, like fences in a garden. You want to find two walls that can hold the most water between them if it rains. Doing this by checking every possible pair of walls manually is like measuring every pair of fences one by one, which takes a lot of time.

The Problem

Manually checking every pair means a lot of repeated work and takes too long, especially if there are many walls. It's easy to make mistakes counting or comparing, and it's very slow, making it frustrating and inefficient.

The Solution

The "Container With Most Water" approach uses two pointers starting at both ends of the row. By moving the pointer at the shorter wall inward, it smartly skips unnecessary checks and quickly finds the best pair. This saves time and effort, making the problem easy to solve.

Before vs After

✗ Before

for i in 0..n {
    for j in i+1..n {
        let height = std::cmp::min(heights[i], heights[j]);
        let width = j - i;
        max_area = std::cmp::max(max_area, height * width);
    }
}

✓ After

let mut left = 0;
let mut right = n - 1;
while left < right {
    let height = std::cmp::min(heights[left], heights[right]);
    max_area = std::cmp::max(max_area, height * (right - left));
    if heights[left] < heights[right] {
        left += 1;
    } else {
        right -= 1;
    }
}

What It Enables

This method lets you quickly find the largest container to hold water without checking every pair, saving time and effort.

Real Life Example

Think of two walls holding water in a garden after rain. Finding the best walls to hold the most water helps in designing efficient water storage or drainage systems.

Key Takeaways

Manual checking is slow and error-prone.

Two-pointer technique smartly narrows down the search.

Efficiently finds the maximum water container in linear time.

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