Trapping Rain Water Problem in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Trapping Rain Water Problem?

To find how much water can be trapped between bars of different heights after raining.

intermediate

What two arrays are commonly used to solve the Trapping Rain Water Problem efficiently?

Left max array and right max array, which store the highest bar from the left and right sides respectively for each position.

beginner

In the Trapping Rain Water Problem, how do you calculate trapped water at a given index?

Trapped water = min(left_max[i], right_max[i]) - height[i], if this value is positive.

intermediate

What is the time complexity of the efficient solution using left max and right max arrays?

O(n), where n is the number of bars, because we traverse the array a few times linearly.

advanced

What is a two-pointer approach in the Trapping Rain Water Problem?

A method using two pointers starting from both ends moving inward, keeping track of max heights to calculate trapped water without extra arrays.

What does the left max array represent in the Trapping Rain Water Problem?

A The lowest bar from the left up to current index

B The highest bar from the left up to current index

C The highest bar from the right up to current index

D The height of the current bar

The left max array stores the tallest bar seen from the left side up to each index.

How do you find the water trapped at index i?

A min(left_max[i], right_max[i]) - height[i]

B height[i] - min(left_max[i], right_max[i])

C left_max[i] + right_max[i] - height[i]

D max(left_max[i], right_max[i]) - height[i]

Water trapped is the minimum of left and right max heights minus the current height.

What is the space complexity of the solution using left max and right max arrays?

A O(1)

B O(n^2)

C O(n)

D O(log n)

We use two extra arrays of size n, so space complexity is O(n).

Which approach can reduce space complexity to O(1) for this problem?

A Using sorting

B Using a stack

C Using recursion

D Using two pointers

The two-pointer approach calculates trapped water on the fly without extra arrays.

What is the main reason trapped water depends on the minimum of left max and right max heights?

A Water spills over the shorter side

B Water fills up to the taller side

C Water fills up to the average height

D Water depends only on the current bar height

Water can only be trapped up to the height of the shorter boundary on either side.

Explain how to solve the Trapping Rain Water Problem using the left max and right max arrays.

Think about how the highest bars on each side limit water trapping.

Describe the two-pointer approach for the Trapping Rain Water Problem and how it optimizes space.

Focus on how pointers move and how water is calculated step-by-step.

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