Trapping Rain Water Using Stack in DSA Java - Build from Scratch

Trapping Rain Water Using Stack

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

🎯 Goal: You will write a Java program that uses a stack to calculate the total amount of trapped rain water given an array of building heights.

📋 What You'll Learn

Create an integer array called heights with the exact values: {0,1,0,2,1,0,1,3,2,1,2,1}

Create an empty stack of integers called stack to store indices

Use a while loop to iterate through the heights array using an index variable i

Inside the loop, use a while loop to pop from stack when current height is greater than height at top of stack

Calculate trapped water using the distance between current index and new top of stack and the bounded height

Print the total trapped water using System.out.println(totalWater);

💡 Why This Matters

🌍 Real World

Calculating trapped rain water is useful in civil engineering and urban planning to understand water accumulation between buildings or terrains.

💼 Career

This problem teaches stack usage and problem-solving skills useful for software engineering interviews and real-world algorithmic challenges.

1

Create the heights array

Create an integer array called heights with these exact values: {0,1,0,2,1,0,1,3,2,1,2,1}

DSA Java

# Your code here

Hint

Use int[] heights = {0,1,0,2,1,0,1,3,2,1,2,1}; to create the array.

2

Create the stack and helper variables

Create an empty Stack<Integer> called stack, an integer i set to 0, and an integer totalWater set to 0

DSA Java

public class TrappingRainWater {
    public static void main(String[] args) {
        int[] heights = {0,1,0,2,1,0,1,3,2,1,2,1};
        // Your code here
    }
}

Hint

Use Stack<Integer> stack = new Stack<>(); to create the stack.

3

Implement the trapping water logic using stack

Use a while loop with condition i < heights.length. Inside it, use another while loop with condition !stack.isEmpty() && heights[i] > heights[stack.peek()]. Inside inner loop, pop from stack into top. If stack is empty after pop, break. Calculate distance = i - stack.peek() - 1 and boundedHeight = Math.min(heights[i], heights[stack.peek()]) - heights[top]. Add distance * boundedHeight to totalWater. After inner loop, push i to stack and increment i by 1.

DSA Java

import java.util.Stack;

public class TrappingRainWater {
    public static void main(String[] args) {
        int[] heights = {0,1,0,2,1,0,1,3,2,1,2,1};
        Stack stack = new Stack();
        int i = 0;
        int totalWater = 0;
        // Your code here
    }
}

Hint

Use nested while loops and stack operations as described to calculate trapped water.

4

Print the total trapped water

Print the value of totalWater using System.out.println(totalWater);

DSA Java

import java.util.Stack;

public class TrappingRainWater {
    public static void main(String[] args) {
        int[] heights = {0,1,0,2,1,0,1,3,2,1,2,1};
        Stack stack = new Stack();
        int i = 0;
        int totalWater = 0;

        while (i  heights[stack.peek()]) {
                int top = stack.pop();
                if (stack.isEmpty()) {
                    break;
                }
                int distance = i - stack.peek() - 1;
                int boundedHeight = Math.min(heights[i], heights[stack.peek()]) - heights[top];
                totalWater += distance * boundedHeight;
            }
            stack.push(i);
            i++;
        }
        // Your code here
    }
}

Hint

Use System.out.println(totalWater); to print the result.

Practice

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

easy

A. To keep track of the indices of bars that are not yet processed for water trapping

B. To store the heights of the bars in sorted order

C. To reverse the array of bar heights

D. To count the total number of 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 how the stack helps in trapping water

   When a bar taller than the top of the stack is found, it means water can be trapped between bars. The stack helps find the left boundary and calculate trapped water.

3. Final Answer:

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

4. Quick Check:

   Stack stores indices for trapping water = C [OK]

Hint: Stack holds indices to find boundaries for trapped water [OK]

Common Mistakes:

• Thinking stack stores heights instead of indices
• Assuming stack sorts bars by height
• Confusing stack use with reversing the array

2. Which of the following Java code snippets correctly initializes an empty stack to be used for trapping rain water indices?

easy

A. Stack stack = new Stack<int>();

B. Stack<Integer> stack = new Stack<>();

C. Stack<int> stack = new Stack<>();

D. Stack<Integer> stack = new Stack<int>();

Solution

1. Step 1: Check Java generic syntax for Stack

   Java uses generics with angle brackets and wrapper classes like Integer, not primitive int.

2. Step 2: Identify correct initialization

   Correct syntax is Stack<Integer> stack = new Stack<>(); which uses Integer wrapper and diamond operator.

3. Final Answer:

   Stack<Integer> stack = new Stack<>(); -> Option B

4. Quick Check:

   Use wrapper Integer and diamond operator = A [OK]

Hint: Use Stack<Integer> with diamond operator for empty stack [OK]

Common Mistakes:

• Using primitive int inside generics
• Omitting generic type causing raw type warning
• Incorrect constructor syntax with generics

3. What is the output of the following Java code snippet that calculates trapped rain water using stack?

int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
System.out.println(trap(height));

public static int trap(int[] height) {
    Stack<Integer> stack = new Stack<>();
    int water = 0, current = 0;
    while (current < height.length) {
        while (!stack.isEmpty() && height[current] > height[stack.peek()]) {
            int top = stack.pop();
            if (stack.isEmpty()) break;
            int distance = current - stack.peek() - 1;
            int boundedHeight = Math.min(height[current], height[stack.peek()]) - height[top];
            water += distance * boundedHeight;
        }
        stack.push(current++);
    }
    return water;
}

medium

A. 5

B. 7

C. 6

D. 8

Solution

1. Step 1: Trace the stack and water calculation

   The code uses a stack to find boundaries and calculates trapped water step-by-step for the given heights.

2. Step 2: Calculate trapped water for given input

   For the input array, the total trapped water calculated is 6 units after processing all bars.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Trapped water for given heights = 6 [OK]

Hint: Trace stack pops and water increments carefully [OK]

Common Mistakes:

• Miscounting distance between bars
• Incorrect bounded height calculation
• Forgetting to break when stack is empty

4. Identify the error in the following Java code snippet for trapping rain water using stack and select the fix.

public static int trap(int[] height) {
    Stack<Integer> stack = new Stack<>();
    int water = 0, current = 0;
    while (current <= height.length) {
        while (!stack.isEmpty() && height[current] > height[stack.peek()]) {
            int top = stack.pop();
            if (stack.isEmpty()) break;
            int distance = current - stack.peek() - 1;
            int boundedHeight = Math.min(height[current], height[stack.peek()]) - height[top];
            water += distance * boundedHeight;
        }
        stack.push(current++);
    }
    return water;
}

medium

A. Change 'stack.pop()' to 'stack.peek()' to avoid removing elements

B. Replace Stack<Integer> with ArrayList<Integer> for better performance

C. Remove the inner while loop to simplify the code

D. Change 'current <= height.length' to 'current < height.length' to avoid ArrayIndexOutOfBoundsException

Solution

1. Step 1: Identify the loop boundary error

   The condition 'current <= height.length' causes an attempt to access height[height.length], which is out of bounds.

2. Step 2: Fix the loop condition

   Changing it to 'current < height.length' prevents the out-of-bounds error and allows correct iteration.

3. Final Answer:

   Change 'current <= height.length' to 'current < height.length' to avoid ArrayIndexOutOfBoundsException -> Option D

4. Quick Check:

   Loop boundary must be less than array length = A [OK]

Hint: Use '<' for array index loops to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Replacing stack with wrong data structure
• Removing necessary inner loop

5. Given the array height = {4, 2, 0, 3, 2, 5}, what is the total amount of trapped rain water calculated by the stack-based algorithm?

hard

A. 9

B. 7

C. 10

D. 8

Solution

1. Step 1: Analyze trapped water between bars

   Between bars of height 4 and 5, water is trapped over lower bars: 2, 0, 3, 2.

2. Step 2: Calculate trapped water volume

   Water level is min(4,5)=4. Water trapped = (4-2) + (4-0) + (4-3) + (4-2) = 2 + 4 + 1 + 2 = 9 units.

3. Final Answer:

   9 -> Option A

4. Quick Check:

   Sum trapped water between boundaries = 9 [OK]

Hint: Calculate water trapped between boundaries using min height [OK]

Common Mistakes:

• Using left boundary height only
• Ignoring right boundary taller bar
• Adding heights instead of differences