Trapping Rain Water Using Stack in DSA Java - Time & Space Complexity

Time Complexity: Trapping Rain Water Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time needed to calculate trapped rain water changes as the input size grows.

Specifically, how does the number of operations grow when using a stack to solve this problem?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int trap(int[] height) {
    int water = 0, current = 0;
    Stack<Integer> stack = new Stack<>();
    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;
}
    

This code calculates how much rain water can be trapped between bars of different heights using a stack to track boundaries.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer while loop runs through each bar once, and the inner while loop pops bars from the stack when the current bar is taller.
• How many times: Each bar is pushed and popped at most once, so total operations related to stack are proportional to the number of bars.

How Execution Grows With Input

As the number of bars increases, each bar is processed a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 (push and pop each bar once)
100	About 200
1000	About 2000

Pattern observation: The operations grow roughly in a straight line with input size, doubling the bars roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute trapped water grows linearly with the number of bars.

Common Mistake

[X] Wrong: "The inner while loop makes this algorithm O(n²) because it can run many times inside the outer loop."

[OK] Correct: Each bar is pushed and popped only once, so total inner loop executions across the whole run cannot exceed n, keeping the total operations linear.

Interview Connect

Understanding this linear time solution shows you can use stacks smartly to solve tricky problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if we replaced the stack with a simple nested loop approach? How would the time complexity change?"

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