Trapping Rain Water Using Stack in DSA Javascript - 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.

How does the number of steps grow when we use a stack to solve this problem?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function trap(height) {
  let stack = [];
  let water = 0;
  for (let i = 0; i < height.length; i++) {
    while (stack.length && height[i] > height[stack[stack.length - 1]]) {
      let top = stack.pop();
      if (!stack.length) break;
      let distance = i - stack[stack.length - 1] - 1;
      let boundedHeight = Math.min(height[i], height[stack[stack.length - 1]]) - height[top];
      water += distance * boundedHeight;
    }
    stack.push(i);
  }
  return water;
}
    

This code calculates how much rainwater can be trapped between bars using a stack to track boundaries.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs once over all bars.
• Nested operation: The while-loop pops from the stack when current bar is higher than the top of stack.
• How many times: Each bar is pushed and popped at most once, so total stack operations are proportional to input size.

How Execution Grows With Input

As the number of bars increases, the for-loop runs once per bar, and the while-loop pops bars only when needed.

Input Size (n)	Approx. Operations
10	About 20 stack operations
100	About 200 stack operations
1000	About 2000 stack operations

Pattern observation: The total operations grow roughly in a straight line with input size.

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: "Because there is a nested while-loop inside the for-loop, the time complexity must be O(n²)."

[OK] Correct: Each bar is pushed and popped only once, so the total number of stack operations is at most 2n, making the overall time linear, not quadratic.

Interview Connect

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

Self-Check

"What if we used a simple nested loop without a stack? 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 store the total amount of trapped water directly

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 array of bar heights

Solution

1. Step 1: Understand the role of stack in the algorithm

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

2. Step 2: Clarify what the stack does not do

   The stack does not store water amounts or sort bars; it helps find boundaries for trapped water.

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 [OK]

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

Common Mistakes:

• Thinking stack stores water volume directly
• Assuming stack sorts the bars
• Believing stack reverses the array

2. Which of the following is the correct way to initialize an empty stack in JavaScript for the trapping rain water problem?

easy

A. const stack = [];

B. const stack = {};

C. const stack = null;

D. const stack = new Stack();

Solution

1. Step 1: Identify the correct data structure for stack in JavaScript

   JavaScript uses arrays as stacks by pushing and popping elements.

2. Step 2: Check each option

   const stack = []; initializes an empty array, which works as a stack. const stack = {}; is an object, not a stack. const stack = null; is null, no stack. const stack = new Stack(); assumes a Stack class which is not built-in.

3. Final Answer:

   const stack = []; -> Option A

4. Quick Check:

   Empty array for stack [OK]

Hint: Use empty array [] as stack in JavaScript [OK]

Common Mistakes:

• Using object {} instead of array
• Assuming Stack class exists by default
• Initializing stack as null

3. What will be the output of the following code snippet that uses stack to calculate trapped rain water?

const height = [0,1,0,2,1,0,1,3,2,1,2,1];
function trap(height) {
  let stack = [];
  let water = 0;
  for (let i = 0; i < height.length; i++) {
    while (stack.length && height[i] > height[stack[stack.length - 1]]) {
      let top = stack.pop();
      if (!stack.length) break;
      let distance = i - stack[stack.length - 1] - 1;
      let boundedHeight = Math.min(height[i], height[stack[stack.length - 1]]) - height[top];
      water += distance * boundedHeight;
    }
    stack.push(i);
  }
  return water;
}
console.log(trap(height));

medium

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand the input and algorithm flow

   The input array represents bar heights. The stack-based algorithm calculates trapped water by finding boundaries and distances.

2. Step 2: Trace the code logic mentally

   The code correctly calculates trapped water for the given heights, which is a classic example resulting in 6 units of water trapped.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Classic trapping rain water example = 6 [OK]

Hint: Classic example height array traps 6 units water [OK]

Common Mistakes:

• Miscounting trapped water units
• Ignoring stack empty check
• Wrong distance or bounded height calculation

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

function trap(height) {
  let stack = [];
  let water = 0;
  for (let i = 0; i < height.length; i++) {
    while (stack.length && height[i] >= height[stack[stack.length - 1]]) {
      let top = stack.pop();
      if (!stack.length) break;
      let distance = i - stack[stack.length - 1] - 1;
      let boundedHeight = Math.min(height[i], height[stack[stack.length - 1]]) - height[top];
      water += distance * boundedHeight;
    }
    stack.push(i);
  }
  return water;
}

medium

A. The distance calculation is incorrect

B. The stack should be initialized as an object, not array

C. The comparison should be >, not >= in the while condition

D. The water variable should be reset inside the loop

Solution

1. Step 1: Analyze the while loop condition

   The condition uses >= which causes incorrect popping when heights are equal, leading to wrong trapped water calculation.

2. Step 2: Correct the condition

   Changing >= to > ensures only strictly higher bars trigger popping, preserving correct boundaries.

3. Final Answer:

   The comparison should be >, not >= in the while condition -> Option C

4. Quick Check:

   Use > in while condition to avoid wrong pops [OK]

Hint: Use >, not >=, to avoid popping equal height bars [OK]

Common Mistakes:

• Using >= causes incorrect water calculation
• Initializing stack as object instead of array
• Resetting water inside loop wrongly

5. Given the heights array [4, 2, 0, 3, 2, 5], what is the total trapped rain water calculated using the stack method?

hard

A. 6

B. 7

C. 8

D. 9

Solution

1. Step 1: Visualize the bars and trapped water

   Bars heights: 4,2,0,3,2,5. Water trapped between bars is calculated by stack method considering boundaries and distances.

2. Step 2: Calculate trapped water step-by-step

   Water at index 1: 2, index 2: 4, index 3: 1, index 4: 2, total 9 units.

3. Final Answer:

   9 -> Option D

4. Quick Check:

   Sum trapped water units = 9 [OK]

Hint: Sum water between boundaries carefully for correct total [OK]

Common Mistakes:

• Miscounting trapped water units
• Ignoring zero height bars
• Wrong boundary identification