Trapping Rain Water Problem in DSA C - Time & Space Complexity

Time Complexity: Trapping Rain Water Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to solve the Trapping Rain Water problem changes as the input size grows.

Specifically, how the number of steps grows when we have more bars in the elevation map.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int trap(int* height, int heightSize) {
    int left = 0, right = heightSize - 1;
    int leftMax = 0, rightMax = 0;
    int water = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= leftMax) leftMax = height[left];
            else water += leftMax - height[left];
            left++;
        } else {
            if (height[right] >= rightMax) rightMax = height[right];
            else water += rightMax - height[right];
            right--;
        }
    }
    return water;
}
    

This code calculates how much water can be trapped between bars represented by an array of heights.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single while loop that moves two pointers from both ends towards the center.
• How many times: The loop runs until the two pointers meet, so roughly once per element in the array.

How Execution Grows With Input

As the number of bars increases, the loop runs once for each bar, so the steps grow directly with input size.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The number of steps grows linearly as input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to solve the problem grows in direct proportion to the number of bars.

Common Mistake

[X] Wrong: "Because there are two pointers, the time is O(n²)."

[OK] Correct: Each pointer moves only forward and never goes back, so together they make at most n steps total, not n times n.

Interview Connect

Understanding this linear time solution shows you can optimize problems by clever pointer use instead of nested loops, a valuable skill in interviews.

Self-Check

"What if we used nested loops to check every pair of bars? How would the time complexity change?"

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