Trapping Rain Water Using Stack in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize an empty stack for indices.

DSA Rust

let mut stack: Vec<usize> = [1];

Drag options to blanks, or click blank then click option'

AVec::new()

Bvec![]

CVec::with_capacity(10)

DVec::default()

Attempts:

3 left

💡 Hint

Common Mistakes

Using vec![] works but is less explicit in this context.

Using Vec::with_capacity(10) preallocates but is not necessary here.

Vec::new() creates a new empty vector which we use as a stack to store indices.

2fill in blank

medium

Complete the code to pop the top index from the stack safely.

DSA Rust

if let Some(top) = stack.[1]() {

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Atake

Bremove

Cpop_back

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

remove() requires an index and is not used for stack pop.

pop_back() does not exist for Vec in Rust.

pop() removes and returns the last element from the vector, which is the top of the stack.

3fill in blank

hard

Fix the error in the condition to check if current height is greater than height at stack top.

DSA Rust

while !stack.is_empty() && height[i] > height[[1]] {

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Astack[stack.len()]

Bstack[stack.len() - 1]

Cstack.last()

Dstack.first()

Attempts:

3 left

💡 Hint

Common Mistakes

stack.last() returns an Option, not an index.

stack[stack.len()] is out of bounds and causes panic.

stack[stack.len() - 1] accesses the last element index in the stack correctly.

4fill in blank

hard

Fill both blanks to calculate the distance and bounded height for trapped water.

DSA Rust

let distance = i - [1] - 1;
let bounded_height = std::cmp::min(height[i], height[[2]]) - height[top];

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Astack[stack.len() - 1]

Bstack.last().copied().unwrap()

Cstack[stack.len() - 2]

Dstack[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using stack[stack.len() - 2] for distance accesses the index before the new top.

Using stack[0] does not represent the new top.

Distance is between current index and the new top after popping (stack[stack.len() - 1]). Bounded height uses min of current height and height at new top (stack.last().copied().unwrap()).

5fill in blank

hard

Fill all three blanks to push current index and update total trapped water.

DSA Rust

stack.[1](i);
total_water += distance * [2];
let top = [3];

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Apush

Bbounded_height

Cstack.pop().unwrap()

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add index causes logic error.

Not unwrapping pop() causes type mismatch.

We push current index i onto stack, add trapped water by multiplying distance and bounded_height, and pop top index from stack.

Practice

1. What is the main purpose of using a stack in the Trapping Rain Water Using Stack algorithm?

easy

A. To reverse the order of the bars

B. To store the total amount of trapped water

C. To keep track of bars that are not yet processed for water trapping

D. To sort the heights of bars in ascending order

Solution

1. Step 1: Understand the role of stack in the algorithm

   The stack stores indices of bars that are waiting to find a right boundary to trap water.

2. Step 2: Clarify what the stack does not do

   The stack does not store water amount or sort bars; it helps find boundaries for trapped water.

3. Final Answer:

   To keep track of bars that are not yet processed for water trapping -> Option C

4. Quick Check:

   Stack tracks bars for trapping water = D [OK]

Hint: Stack holds bars waiting for right boundary [OK]

Common Mistakes:

• Thinking stack stores water amount
• Assuming stack sorts the bars
• Believing stack reverses bar order

2. Which of the following Rust code snippets correctly declares an empty stack to use in the trapping rain water algorithm?

easy

A. let stack = Vec::with_capacity(10);

B. let mut stack: Vec<usize> = Vec::new();

C. let stack = vec![0];

D. let mut stack: Vec<i32> = Vec::new();

Solution

1. Step 1: Identify correct stack declaration for indices

   Indices are usize type, so stack must be Vec<usize> and mutable to push/pop.

2. Step 2: Check each option

   let mut stack: Vec<usize> = Vec::new(); correctly declares mutable Vec<usize> empty stack. let stack = Vec::with_capacity(10); is immutable and no mut keyword. let stack = vec![0]; initializes with one element, not empty. let mut stack: Vec<i32> = Vec::new(); uses i32 instead of usize for indices.

3. Final Answer:

   let mut stack: Vec<usize> = Vec::new(); -> Option B

4. Quick Check:

   Mutable Vec<usize> empty stack = A [OK]

Hint: Use mutable Vec<usize> for stack of indices [OK]

Common Mistakes:

• Using wrong type like i32 for indices
• Not making stack mutable
• Initializing stack with elements instead of empty

3. What is the output of the following Rust code snippet that calculates trapped rain water using stack?

fn trap(height: Vec<i32>) -> i32 {
    let mut stack: Vec<usize> = Vec::new();
    let mut water = 0;
    for i in 0..height.len() {
        while !stack.is_empty() && height[i] > height[*stack.last().unwrap()] {
            let top = stack.pop().unwrap();
            if stack.is_empty() {
                break;
            }
            let distance = i - *stack.last().unwrap() - 1;
            let bounded_height = height[i].min(height[*stack.last().unwrap()]) - height[top];
            water += (distance as i32) * bounded_height;
        }
        stack.push(i);
    }
    water
}

let heights = vec![0,1,0,2,1,0,1,3,2,1,2,1];
println!("{}", trap(heights));

medium

A. 6

B. 5

C. 4

D. 7

Solution

1. Step 1: Understand the input and algorithm

   The input heights represent bars: [0,1,0,2,1,0,1,3,2,1,2,1]. The stack-based algorithm calculates trapped water by finding bounded areas.

2. Step 2: Calculate trapped water step-by-step

   The known trapped water for this input is 6 units, as the algorithm sums water trapped between bars correctly.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Trapped water for given heights = 6 [OK]

Hint: Known test case result is 6 units trapped water [OK]

Common Mistakes:

• Miscounting trapped water units
• Ignoring boundary conditions
• Incorrectly calculating distance or height

4. Identify the bug in the following Rust code snippet for trapping rain water using stack and select the fix:

fn trap(height: Vec<i32>) -> i32 {
    let mut stack: Vec<usize> = Vec::new();
    let mut water = 0;
    for i in 0..height.len() {
        while !stack.is_empty() && height[i] > height[*stack.last().unwrap()] {
            let top = stack.pop().unwrap();
            if stack.is_empty() {
                break;
            }
            let distance = i - *stack.last().unwrap() - 1;
            let bounded_height = height[i].min(height[*stack.last().unwrap()]) - height[top];
            water += distance * bounded_height;
        }
        stack.push(i);
    }
    water
}

medium

A. Change height[i] > height[stack.last().unwrap()] to height[i] >= height[stack.last().unwrap()]

B. Change stack.pop().unwrap() to stack.pop();

C. Remove the if stack.is_empty() { break; } check

D. Change water += distance * bounded_height; to water += (distance as i32) * bounded_height;

Solution

1. Step 1: Identify type mismatch in multiplication

   distance is usize, bounded_height is i32; multiplying usize * i32 causes type error in Rust.

2. Step 2: Fix by casting distance to i32 before multiplication

   Cast distance as i32: (distance as i32) * bounded_height to fix type mismatch.

3. Final Answer:

   Change water += distance * bounded_height; to water += (distance as i32) * bounded_height; -> Option D

4. Quick Check:

   Cast usize to i32 before multiply = C [OK]

Hint: Cast usize to i32 before multiplying [OK]

Common Mistakes:

• Ignoring type mismatch errors
• Removing necessary empty stack check
• Changing comparison operator incorrectly

5. Given the heights array [4, 2, 0, 3, 2, 5], what is the total trapped rain water calculated by the stack-based algorithm?

hard

A. 9

B. 8

C. 6

D. 7

Solution

1. Step 1: Visualize the bars and trapped water

   Bars heights: 4, 2, 0, 3, 2, 5. Water trapped between bars is calculated by stack algorithm using boundaries.

2. Step 2: Calculate trapped water stepwise

   Using min(left_max, right_max) - height[i] for each i:
   - i=1: min(4,5)-2=2
   - i=2: min(4,5)-0=4
   - i=3: min(4,5)-3=1
   - i=4: min(4,5)-2=2
   Total trapped water = 9 units.

3. Final Answer:

   9 -> Option A

4. Quick Check:

   Sum trapped water units = 9 [OK]

Hint: Sum water trapped between boundaries carefully [OK]

Common Mistakes:

• Miscounting water units between bars
• Ignoring zero height bars
• Not considering both left and right boundaries