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, we ask: How does the number of steps grow when the height array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int trap(vector<int>& height) {
    int n = height.size();
    vector<int> leftMax(n), rightMax(n);
    leftMax[0] = height[0];
    for (int i = 1; i < n; i++) {
        leftMax[i] = max(leftMax[i-1], height[i]);
    }
    rightMax[n-1] = height[n-1];
    for (int i = n-2; i >= 0; i--) {
        rightMax[i] = max(rightMax[i+1], height[i]);
    }
    int water = 0;
    for (int i = 0; i < n; i++) {
        water += min(leftMax[i], rightMax[i]) - height[i];
    }
    return water;
}
    

This code calculates how much water can be trapped after raining given an array of heights.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three separate loops each running through the height array once.
• How many times: Each loop runs exactly n times, where n is the size of the height array.

How Execution Grows With Input

As the input size n grows, the number of operations grows roughly in a straight line with n.

Input Size (n)	Approx. Operations
10	About 30 steps (3 loops x 10)
100	About 300 steps (3 loops x 100)
1000	About 3000 steps (3 loops x 1000)

Pattern observation: The steps increase directly with input size, tripled due to three loops but still linear.

Final Time Complexity

Time Complexity: O(n)

This means the time to solve the problem grows in a straight line with the size of the input array.

Common Mistake

[X] Wrong: "Because there are nested max and min operations, the time complexity is quadratic O(n²)."

[OK] Correct: Each max or min is done inside a single loop, not nested loops. So the operations are done one after another, not multiplied.

Interview Connect

Understanding this linear time solution shows you can break down a problem into simple passes over data, a key skill in many coding challenges.

Self-Check

"What if we tried to find trapped water using nested loops checking every pair of bars? How would the time complexity change?"

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 gives an array representing heights of bars and asks how much water can be trapped between them after it rains.

2. Step 2: Identify the goal

   The goal is to calculate the total trapped water volume, not just heights or counts.

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 gaps [OK]

Common Mistakes:

• Confusing trapped water with highest bar height
• Summing bar heights instead of trapped water
• Counting bars taller than a certain height

2. Which of the following C++ code snippets correctly initializes a vector named heights with values {0,1,0,2,1,0,1,3,2,1,2,1}?

easy

A. std::vector<int> heights = (0,1,0,2,1,0,1,3,2,1,2,1);

B. std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1};

C. vector<int> heights(0,1,0,2,1,0,1,3,2,1,2,1);

D. std::vector heights = {0,1,0,2,1,0,1,3,2,1,2,1};

Solution

1. Step 1: Check vector initialization syntax

   In C++, std::vector<int> can be initialized with curly braces and values inside, like {0,1,...}.

2. Step 2: Identify correct syntax

   std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1}; uses correct syntax with std::vector<int> and brace initialization.

3. Final Answer:

   std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1}; -> Option B

4. Quick Check:

   Correct vector init uses braces and type [OK]

Hint: Use curly braces {} for vector initialization [OK]

Common Mistakes:

• Omitting std:: before vector
• Using parentheses instead of braces
• Not specifying vector type

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

std::vector<int> height = {0,1,0,2,1,0,1,3,2,1,2,1};
int n = height.size();
int left_max = 0, right_max = 0, left = 0, right = n - 1, 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++;
  } else {
    if (height[right] >= right_max) right_max = height[right];
    else water += right_max - height[right];
    right--;
  }
}
std::cout << water << std::endl;

medium

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand two-pointer approach

   Two pointers move inward, tracking max heights on left and right, accumulating trapped water where current height is less than max.

2. Step 2: Trace the code logic on given input

   For input {0,1,0,2,1,0,1,3,2,1,2,1}, the total trapped water calculated is 6 units.

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 or right_max correctly
• Adding negative water amounts
• Confusing left and right pointer moves

4. Identify the bug in this C++ code snippet for trapping rain water and select the fix:

std::vector<int> height = {4,2,0,3,2,5};
int n = height.size();
int left_max = 0, right_max = 0, left = 0, right = n - 1, 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++;
  } else {
    if (height[right] > right_max) right_max = height[right];
    else water += right_max - height[right];
    right--;
  }
}
std::cout << water << std::endl;

medium

A. Initialize left_max and right_max to height[0] and height[n-1] respectively

B. Use '>=' instead of '>' when updating left_max and right_max

C. Change while condition to 'left < right' to avoid double counting

D. No bug, code is correct

Solution

1. Step 1: Analyze loop condition impact

   The loop uses 'left <= right', which causes the middle element to be processed twice, leading to incorrect water calculation.

2. Step 2: Fix loop condition

   Changing condition to 'left < right' ensures each element is processed once, preventing double counting.

3. Final Answer:

   Change while condition to 'left < right' to avoid double counting -> Option C

4. Quick Check:

   Loop condition 'left < right' fixes double count bug [OK]

Hint: Use strict less than in two-pointer loops [OK]

Common Mistakes:

• Using <= instead of < in while loop
• Not updating max heights correctly
• Assuming no bug without testing

5. Given an array representing heights of bars, how would you modify the standard trapping rain water algorithm to handle very large input efficiently without extra space?

hard

A. Use recursion to calculate trapped water for each bar

B. Precompute left_max and right_max arrays and sum trapped water using extra arrays

C. Sort the array and calculate trapped water based on sorted positions

D. Use two pointers from both ends, track max heights, and calculate trapped water on the fly

Solution

1. Step 1: Understand space constraints

   Precomputing arrays uses extra space O(n), which is costly for very large inputs.

2. Step 2: Use two-pointer approach for O(1) space

   Two pointers from both ends track max heights and calculate trapped water without extra arrays, making it efficient.

3. Final Answer:

   Use two pointers from both ends, track max heights, and calculate trapped water on the fly -> Option D

4. Quick Check:

   Two-pointer approach = efficient O(1) space [OK]

Hint: Two pointers save space by avoiding extra arrays [OK]

Common Mistakes:

• Using extra arrays causing high memory use
• Sorting changes bar order, invalid for trapping water
• Recursion adds overhead and complexity