Why Trapping Rain Water Using Stack in DSA Java?

The Big Idea

Discover how a simple stack can solve a tricky water-trapping puzzle in just one pass!

The Scenario

Imagine you have a row of buildings with different heights, and after rain, water collects between them. You want to find out how much water is trapped. Trying to measure this by looking at each building and guessing the water trapped manually is tough and confusing.

The Problem

Manually calculating trapped water means checking every building and comparing it with all others on the left and right to find the highest walls. This is slow and easy to mess up, especially with many buildings. It's like trying to count water drops between uneven walls by eye.

The Solution

Using a stack helps keep track of buildings that can trap water. The stack stores building indexes, and when a taller building comes, it calculates trapped water with the help of the stack. This method is faster and less error-prone because it processes buildings in one pass.

Before vs After

✗ Before

int trappedWater = 0;
for (int i = 1; i < heights.length - 1; i++) {
  int leftMax = 0, rightMax = 0;
  for (int j = 0; j <= i; j++) leftMax = Math.max(leftMax, heights[j]);
  for (int j = i; j < heights.length; j++) rightMax = Math.max(rightMax, heights[j]);
  trappedWater += Math.min(leftMax, rightMax) - heights[i];
}

✓ After

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

What It Enables

This approach lets you quickly and accurately find trapped rainwater between buildings in just one pass, even for large inputs.

Real Life Example

City planners can use this method to estimate how much rainwater will collect between buildings to design better drainage systems and prevent flooding.

Key Takeaways

Manual checking is slow and error-prone for large data.

Stack helps track boundaries and calculate trapped water efficiently.

One-pass solution saves time and reduces mistakes.

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