Trapping Rain Water Using Stack in DSA Rust - Time & Space Complexity

Time Complexity: Trapping Rain Water Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time needed to calculate trapped rain water changes as the input size grows.

Specifically, we ask: How does the number of steps grow when we use a stack to solve this problem?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn trap(height: Vec<i32>) -> i32 {
    let mut stack = 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
}
    

This code calculates how much rain water can be trapped between bars using a stack to track boundaries.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop over the height array with an inner while loop that pops from the stack.
• How many times: Each element is pushed once and popped at most once, so the inner loop runs in total at most n times.

How Execution Grows With Input

As the input size n grows, each element is processed a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 steps (push and pop combined)
100	About 200 steps
1000	About 2000 steps

Pattern observation: The total steps grow roughly in a straight line with input size, not squared or faster.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute trapped water grows linearly with the number of bars.

Common Mistake

[X] Wrong: "The inner while loop makes this algorithm O(n²) because it can run many times for each element."

[OK] Correct: Each element is pushed and popped only once, so the total inner loop runs at most n times, keeping the overall time linear.

Interview Connect

Understanding this linear time solution shows you can use stacks smartly to solve tricky problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if we replaced the stack with a simple nested loop approach? How would the time complexity change?"

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