Trapping Rain Water Problem in DSA C - 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.

beginner

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

It stores the highest bar height from the left up to each position.

intermediate

Why do we need both 'left max' and 'right max' arrays?

Because water trapped at each bar depends on the tallest bars on both left and right sides.

beginner

How is the water trapped at a single bar calculated?

Water trapped = minimum of (left max, right max) - height of the bar, if positive.

intermediate

What is the time complexity of the standard 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.

What does the Trapping Rain Water Problem calculate?

A Maximum height of bars

B Amount of water trapped between bars

C Number of bars

D Total length of bars

The problem calculates how much water can be trapped between bars after raining.

Which arrays are used to store the highest bars from left and right?

A start max and end max

B height and width

C top max and bottom max

D left max and right max

We use left max and right max arrays to store the tallest bars from left and right sides.

How do you calculate water trapped at a bar?

A min(left max, right max) - height

B max(left max, right max) + height

C left max + right max

D height - min(left max, right max)

Water trapped is the minimum of left max and right max minus the bar's height.

What is the time complexity of the standard trapping rain water solution?

A O(n)

B O(n^2)

C O(log n)

D O(1)

The solution runs in linear time O(n) by traversing the array multiple times.

If a bar is the tallest on both sides, how much water can it trap?

A Depends on neighbors

B Maximum possible

C Zero

D Equal to its height

A tallest bar cannot trap water on top of itself, so trapped water is zero.

Explain how to calculate the total trapped rain water given an array of bar heights.

Think about the tallest bars on both sides of each bar.

Describe the role of the left max and right max arrays in the Trapping Rain Water Problem.

They help find the boundaries for water trapping.

Practice

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

easy

A. The minimum height of the bars

B. The maximum height of the bars

C. The number of bars in the array

D. The total amount of water trapped between the bars after raining

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 the goal

   The goal is to calculate how much water can be trapped after raining between these bars.

3. Final Answer:

   The total amount of water trapped between the bars after raining -> Option D

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 bar heights
• Calculating max or min height instead of trapped water
• Ignoring the space between bars

2. Which of the following C code snippets correctly declares an array to store bar heights for the Trapping Rain Water problem?

easy

A. int heights[] = {4, 2, 0, 3, 2, 5};

B. int heights = {4, 2, 0, 3, 2, 5};

C. int heights[6] = (4, 2, 0, 3, 2, 5);

D. int heights[6] = [4, 2, 0, 3, 2, 5];

Solution

1. Step 1: Check array declaration syntax in C

   Arrays are declared with type, name, and initializer list in braces {}.

2. Step 2: Validate each option

   int heights[] = {4, 2, 0, 3, 2, 5}; uses correct syntax: int heights[] = {4, 2, 0, 3, 2, 5}; Options A, B, D have syntax errors.

3. Final Answer:

   int heights[] = {4, 2, 0, 3, 2, 5}; -> Option A

4. Quick Check:

   Correct C array syntax = int heights[] = {4, 2, 0, 3, 2, 5}; [OK]

Hint: Use braces {} to initialize arrays in C [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Missing array size or incorrect initializer syntax
• Using square brackets [] for initialization

3. What is the output of the following C code snippet that calculates trapped water?

int heights[] = {3, 0, 2, 0, 4};
int n = 5;
int left_max[5], right_max[5];
int water = 0;

// Calculate left max
left_max[0] = heights[0];
for (int i = 1; i < n; i++) {
    left_max[i] = (heights[i] > left_max[i-1]) ? heights[i] : left_max[i-1];
}

// Calculate right max
right_max[n-1] = heights[n-1];
for (int i = n-2; i >= 0; i--) {
    right_max[i] = (heights[i] > right_max[i+1]) ? heights[i] : right_max[i+1];
}

// Calculate trapped water
for (int i = 0; i < n; i++) {
    water += (left_max[i] < right_max[i] ? left_max[i] : right_max[i]) - heights[i];
}

printf("%d\n", water);

medium

A. 4

B. 6

C. 7

D. 5

Solution

1. Step 1: Calculate left_max and right_max arrays

   left_max = [3, 3, 3, 3, 4], right_max = [4, 4, 4, 4, 4]

2. Step 2: Calculate trapped water at each index

   Water trapped = sum of min(left_max[i], right_max[i]) - heights[i] = (3-3)+(3-0)+(3-2)+(3-0)+(4-4) = 0+3+1+3+0 = 7

3. Final Answer:

   7 -> Option C

4. Quick Check:

   Sum trapped water = 7 [OK]

Hint: Use left_max and right_max arrays to find trapped water fast [OK]

Common Mistakes:

• Mixing up left_max and right_max arrays
• Forgetting to subtract heights from min of max arrays
• Off-by-one errors in loops

4. Identify the bug in this C code snippet for calculating trapped rain water:

int trap(int* height, int heightSize) {
    int left_max[heightSize];
    int right_max[heightSize];
    int water = 0;

    left_max[0] = height[0];
    for (int i = 1; i < heightSize; i++) {
        left_max[i] = height[i] > left_max[i-1] ? height[i] : left_max[i-1];
    }

    right_max[heightSize-1] = height[heightSize-1];
    for (int i = heightSize-2; i >= 0; i--) {
        right_max[i] = height[i] > right_max[i+1] ? height[i] : right_max[i+1];
    }

    for (int i = 0; i < heightSize; i++) {
        water += (left_max[i] < right_max[i] ? left_max[i] : right_max[i]) - height[i];
    }

    return water;
}

medium

A. Incorrect comparison operator in left_max calculation

B. Using variable length arrays (VLA) which may not be supported

C. Wrong loop range in right_max calculation

D. Missing initialization of water variable

Solution

1. Step 1: Analyze array declarations

   The code uses int left_max[heightSize]; and int right_max[heightSize]; which are variable length arrays (VLA).

2. Step 2: Understand VLA support in C

   VLAs are optional in C11 and may not be supported by all compilers, causing portability issues or errors.

3. Final Answer:

   Using variable length arrays (VLA) which may not be supported -> Option B

4. Quick Check:

   VLAs may cause errors = Using variable length arrays (VLA) which may not be supported [OK]

Hint: Avoid VLAs; use dynamic memory or fixed size arrays [OK]

Common Mistakes:

• Assuming VLAs always work in all C compilers
• Ignoring compiler warnings about VLAs
• Confusing VLA issue with logic errors

5. Given the heights array {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1}, what is the total trapped rain water using the two-pointer approach in C?

hard

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand two-pointer approach

   Use two pointers left and right starting at ends, track left_max and right_max, move pointers inward calculating trapped water.

2. Step 2: Dry run the array

   Heights: [0,1,0,2,1,0,1,3,2,1,2,1] Water trapped = 6 units after calculating min(left_max, right_max) - height at each step.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Two-pointer trapped water = 6 [OK]

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

Common Mistakes:

• Not updating left_max and right_max correctly
• Moving pointers incorrectly causing missed water
• Confusing two-pointer with prefix arrays method