Trapping Rain Water Problem in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the total water trapped variable.

DSA Rust

let mut total_water = [1];

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

A0

B1

Cheight.len()

Dusize::MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing total_water to 1 or a large number.

We start counting trapped water from zero because initially no water is trapped.

2fill in blank

medium

Complete the code to update the left maximum height while iterating.

DSA Rust

left_max = [1](left_max, height[left]);

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Astd::cmp::min

Bstd::cmp::sum

Cstd::cmp::abs

Dstd::cmp::max

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max causes incorrect water calculation.

We update left_max to the maximum height seen so far on the left side.

3fill in blank

hard

Fix the error in the condition to decide which pointer to move.

DSA Rust

if height[left] < [1] {

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Aheight[right]

Bleft_max

Ctotal_water

Dright_max

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with left_max, right_max, or total_water causes wrong logic.

We compare height[left] with height[right] to decide pointer movement.

4fill in blank

hard

Fill both blanks to correctly calculate trapped water and move pointers.

DSA Rust

if height[left] < height[right] {
    total_water += [1] - height[left];
    left += [2];
} else {
    total_water += right_max - height[right];
    right -= 1;
}

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

Aleft_max

Bright_max

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using right_max on left side or moving pointer by 0.

We add trapped water by subtracting current height from left_max and move left pointer forward by 1.

5fill in blank

hard

Fill all three blanks to complete the trapping rain water function.

DSA Rust

fn trap(height: Vec<i32>) -> i32 {
    let (mut left, mut right) = (0, height.len() - 1);
    let (mut left_max, mut right_max) = (0, 0);
    let mut total_water = 0;
    while left < right {
        if height[left] < height[right] {
            left_max = [1](left_max, height[left]);
            total_water += left_max - height[left];
            left += [2];
        } else {
            right_max = [3](right_max, height[right]);
            total_water += right_max - height[right];
            right -= 1;
        }
    }
    total_water
}

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

Astd::cmp::max

B1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max or moving pointer by 0.

We update left_max and right_max with max function, move left pointer by 1.

Practice

1. What does the Trapping Rain Water problem ask you to find given an array of heights?

easy

A. The total amount of water that can be trapped between the bars after raining

B. The highest bar in the array

C. The number of bars taller than a given height

D. The sum of all bar heights

Solution

1. Step 1: Understand the problem context

   The problem involves an array representing heights of bars where water can be trapped between them.

2. Step 2: Identify what is asked

   We need to find how much water can be trapped after raining, which depends on the heights and gaps between bars.

3. Final Answer:

   The total amount of water that can be trapped between the bars after raining -> Option A

4. Quick Check:

   Trapping Rain Water = Total trapped water [OK]

Hint: Think of bars as walls holding water in between [OK]

Common Mistakes:

• Confusing trapped water with total height sum
• Ignoring gaps between bars
• Only finding tallest bar instead of trapped water

2. Which Rust code snippet correctly initializes a vector of bar heights for the Trapping Rain Water problem?

easy

A. let heights = [0,1,0,2,1,0,1,3,2,1,2,1];

B. let heights = vec![0,1,0,2,1,0,1,3,2,1,2,1];

C. let heights = (0,1,0,2,1,0,1,3,2,1,2,1);

D. let heights = {0,1,0,2,1,0,1,3,2,1,2,1};

Solution

1. Step 1: Identify Rust vector syntax

   In Rust, vectors are created using vec![elements], which is dynamic and growable.

2. Step 2: Check each option

   let heights = vec![0,1,0,2,1,0,1,3,2,1,2,1]; uses vec! macro correctly; B is an array, C is a tuple, D is invalid syntax.

3. Final Answer:

   let heights = vec![0,1,0,2,1,0,1,3,2,1,2,1]; -> Option B

4. Quick Check:

   vec! macro creates vector [OK]

Hint: Use vec![] to create vectors in Rust [OK]

Common Mistakes:

• Using array syntax instead of vector
• Using tuple parentheses instead of brackets
• Using curly braces which are invalid here

3. What is the output of this Rust code snippet that calculates trapped water?

fn trap(height: Vec<i32>) -> i32 {
    let mut left = 0;
    let mut right = height.len() - 1;
    let mut left_max = 0;
    let mut right_max = 0;
    let mut water = 0;
    while left < right {
        if height[left] < height[right] {
            if height[left] > left_max {
                left_max = height[left];
            } else {
                water += left_max - height[left];
            }
            left += 1;
        } else {
            if height[right] > right_max {
                right_max = height[right];
            } else {
                water += right_max - height[right];
            }
            right -= 1;
        }
    }
    water
}

fn main() {
    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 two-pointer approach in code

   The code uses two pointers from left and right, tracking max heights and accumulating trapped water.

2. Step 2: Trace the example input

   For input [0,1,0,2,1,0,1,3,2,1,2,1], the trapped water calculated is 6 units.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Two-pointer trap water = 6 [OK]

Hint: Use two pointers and track max heights from both ends [OK]

Common Mistakes:

• Miscounting trapped water units
• Confusing left_max and right_max updates
• Off-by-one errors in pointer movement

4. Identify the bug in this Rust function for trapping rain water:

fn trap(height: Vec<i32>) -> i32 {
    let mut left = 0;
    let mut right = height.len();
    let mut left_max = 0;
    let mut right_max = 0;
    let mut water = 0;
    while left < right {
        if height[left] < height[right] {
            if height[left] > left_max {
                left_max = height[left];
            } else {
                water += left_max - height[left];
            }
            left += 1;
        } else {
            if height[right] > right_max {
                right_max = height[right];
            } else {
                water += right_max - height[right];
            }
            right -= 1;
        }
    }
    water
}

medium

A. water should be a float instead of i32

B. left_max and right_max should start at 1

C. The while loop condition should be left <= right

D. right should be initialized to height.len() - 1

Solution

1. Step 1: Check initialization of right pointer

   right is set to height.len(), which is out of bounds for indexing (max index is len()-1).

2. Step 2: Correct right initialization

   right should be height.len() - 1 to point to the last valid index.

3. Final Answer:

   right should be initialized to height.len() - 1 -> Option D

4. Quick Check:

   Indexing error fixed by right = len()-1 [OK]

Hint: Array indices go from 0 to len-1, not len [OK]

Common Mistakes:

• Using len() as index causes runtime panic
• Changing loop condition incorrectly
• Starting max heights at wrong values

5. Given the heights array [4,2,0,3,2,5], what is the total trapped rain water using the two-pointer approach?

hard

A. 6

B. 7

C. 9

D. 8

Solution

1. Step 1: Analyze heights and max boundaries

   Left max starts at 4, right max at 5. Water trapped depends on min(left_max, right_max) minus current height.

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

   Positions: - index 1: min(4,5)=4, height=2, trapped=2 - index 2: min(4,5)=4, height=0, trapped=4 - index 3: min(4,5)=4, height=3, trapped=1 - index 4: min(4,5)=4, height=2, trapped=2 Total trapped water = 2+4+1+2 = 9 units. Verified with two-pointer simulation.

3. Final Answer:

   9 -> Option C

4. Quick Check:

   Two-pointer trap water for [4,2,0,3,2,5] = 9 [OK]

Hint: Calculate trapped water by min(left_max, right_max) - height [OK]

Common Mistakes:

• Adding trapped water incorrectly at edges
• Confusing left_max and right_max updates
• Forgetting to move pointers properly