Trapping Rain Water Using Stack in DSA C++ - Build from Scratch

Trapping Rain Water Using Stack

📖 Scenario: Imagine you have a row of walls of different heights. When it rains, water gets trapped between these walls. We want to find out how much water is trapped after the rain stops.

🎯 Goal: You will write a program that uses a stack to calculate the total amount of trapped rainwater given the heights of the walls.

📋 What You'll Learn

Create a vector called height with the exact values: {0,1,0,2,1,0,1,3,2,1,2,1}

Create an empty stack of integers called st

Use a while loop and if conditions to implement the trapping rain water logic using the stack

Print the total trapped water as an integer

💡 Why This Matters

🌍 Real World

Calculating trapped rainwater helps in understanding water flow and storage in terrains and urban planning.

💼 Career

This problem is common in coding interviews and helps develop problem-solving skills using stacks and arrays.

1

Create the height vector

Create a vector of integers called height with these exact values: {0,1,0,2,1,0,1,3,2,1,2,1}

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector height = {0,1,0,2,1,0,1,3,2,1,2,1};
    // Your code here

    return 0;
}

Hint

Use std::vector<int> and initialize it with the given values.

2

Create an empty stack and initialize variables

Create an empty stack of integers called st and an integer variable water initialized to 0

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector height = {0,1,0,2,1,0,1,3,2,1,2,1};
    std::stack st;
    int water = 0;
    // Your code here

    return 0;
}

Hint

Use std::stack<int> st; to create the stack and int water = 0; to start counting trapped water.

3

Implement the trapping rain water logic using stack

Use a for loop with variable i from 0 to height.size(). Inside the loop, use a while loop to check if the stack is not empty and height[i] is greater than height[st.top()]. Pop the top of the stack into top. If the stack becomes empty, break. Calculate the distance as i - st.top() - 1 and bounded height as std::min(height[i], height[st.top()]) - height[top]. Add distance * bounded_height to water. After the while loop, push i onto the stack.

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector height = {0,1,0,2,1,0,1,3,2,1,2,1};
    std::stack st;
    int water = 0;

    for (int i = 0; i < height.size(); i++) {
        // Your code here
    }

    return 0;
}

Hint

Use a for loop to go through each wall. Use a while loop to pop from the stack when current height is greater. Calculate trapped water using distance and bounded height.

4

Print the total trapped water

Print the value of water using std::cout

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::vector height = {0,1,0,2,1,0,1,3,2,1,2,1};
    std::stack st;
    int water = 0;

    for (int i = 0; i  height[st.top()]) {
            int top = st.top();
            st.pop();
            if (st.empty()) {
                break;
            }
            int distance = i - st.top() - 1;
            int bounded_height = std::min(height[i], height[st.top()]) - height[top];
            water += distance * bounded_height;
        }
        st.push(i);
    }

    std::cout << water << std::endl;
}

Hint

Use std::cout << water << std::endl; to print the total trapped water.

Practice

1. What is the main purpose of using a stack in the Trapping Rain Water Using Stack algorithm?

easy

A. To store the total amount of trapped water

B. To keep track of bars that are not yet processed for water trapping

C. To sort the heights of bars in ascending order

D. To reverse the order of the bars

Solution

1. Step 1: Understand the role of the stack

   The stack stores indices of bars that are waiting to find a right boundary to trap water.

2. Step 2: Explain why other options are incorrect

   The stack does not store water amount or sort bars; it helps track bars for water calculation.

3. Final Answer:

   To keep track of bars that are not yet processed for water trapping -> Option B

4. Quick Check:

   Stack tracks bars for trapping water = A [OK]

Hint: Stack holds bars waiting for right boundary [OK]

Common Mistakes:

• Thinking stack stores water amount
• Assuming stack sorts bars
• Confusing stack with array for heights

2. Which of the following C++ code snippets correctly declares a stack to be used for trapping rain water indices?

easy

A. std::stack<int> st;

B. stack<int> st = new stack<int>();

C. std::stack st<int>;

D. stack<int> st();

Solution

1. Step 1: Check correct C++ stack declaration syntax

   Correct syntax is std::stack<int> st; to declare a stack of integers.

2. Step 2: Identify errors in other options

   stack<int> st = new stack<int>(); uses incorrect syntax with 'new'; C has wrong template placement; D declares a function, not a stack.

3. Final Answer:

   std::stack<int> st; -> Option A

4. Quick Check:

   Correct stack declaration = B [OK]

Hint: Use std::stack<int> st; to declare stack [OK]

Common Mistakes:

• Using 'new' keyword incorrectly
• Wrong template syntax
• Declaring function instead of stack

3. What is the output of the following C++ code snippet that uses stack to calculate trapped rain water?

int height[] = {0,1,0,2,1,0,1,3,2,1,2,1};
int n = 12;
std::stack<int> st;
int water = 0;
for (int i = 0; i < n; i++) {
  while (!st.empty() && height[i] > height[st.top()]) {
    int top = st.top();
    st.pop();
    if (st.empty()) break;
    int distance = i - st.top() - 1;
    int bounded_height = std::min(height[i], height[st.top()]) - height[top];
    water += distance * bounded_height;
  }
  st.push(i);
}
std::cout << water << std::endl;

medium

A. 8

B. 5

C. 6

D. 7

Solution

1. Step 1: Trace the stack operations and water calculation

   The code calculates trapped water for the given heights array using stack logic. The total trapped water is 6 units.

2. Step 2: Verify calculations with known example

   This is a classic example where trapped water is 6 units, matching the output of the code.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Trapped water = 6 [OK]

Hint: Classic example traps 6 units water [OK]

Common Mistakes:

• Miscounting distance between bars
• Wrong bounded height calculation
• Ignoring empty stack break condition

4. Identify the error in the following C++ code snippet for trapping rain water using stack:

std::stack<int> st;
int water = 0;
for (int i = 0; i < height.size(); i++) {
  while (!st.empty() && height[i] >= height[st.top()]) {
    int top = st.top();
    st.pop();
    if (st.empty()) break;
    int distance = i - st.top() - 1;
    int bounded_height = std::min(height[i], height[st.top()]) - height[top];
    water += distance * bounded_height;
  }
  st.push(i);
}
std::cout << water << std::endl;

medium

A. No break condition inside while loop

B. Missing initialization of water variable

C. Stack should store heights, not indices

D. Using >= instead of > in while condition causes incorrect water calculation

Solution

1. Step 1: Analyze the while loop condition

   The condition uses >= which causes popping bars of equal height, leading to incorrect trapped water calculation.

2. Step 2: Explain correct condition and impact

   Using > ensures only bars lower than current are popped, preserving correct boundaries for water trapping.

3. Final Answer:

   Using >= instead of > in while condition causes incorrect water calculation -> Option D

4. Quick Check:

   Use > not >= in while condition [OK]

Hint: Use > not >= in while condition [OK]

Common Mistakes:

• Using >= instead of > in while condition
• Storing heights instead of indices in stack
• Forgetting break when stack is empty

5. Given the heights array {4, 2, 0, 3, 2, 5}, what is the total trapped rain water calculated using the stack approach?

hard

A. 8

B. 7

C. 9

D. 6

Solution

1. Step 1: Calculate trapped water between bars using stack logic

   Between bars, trapped water units are: 2 units between 4 and 3, 1 unit between 3 and 5, and 5 units between 2 and 5, totaling 8 units.

2. Step 2: Sum trapped water units

   Sum of trapped water is 8 units for the given array.

3. Final Answer:

   8 -> Option A

4. Quick Check:

   Trapped water total = 8 [OK]

Hint: Sum water trapped between boundaries carefully [OK]

Common Mistakes:

• Ignoring zero height bars
• Miscounting trapped water between bars
• Not considering both left and right boundaries