Pop Operation on Stack in DSA Javascript - Time & Space Complexity
Let's explore how long it takes to remove an item from a stack using the pop operation.
We want to know how the time needed changes as the stack grows.
Analyze the time complexity of the following code snippet.
class Stack {
constructor() {
this.items = [];
}
pop() {
if (this.items.length === 0) return null;
return this.items.pop();
}
}
This code defines a stack and removes the top item when pop is called.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Removing the last element from the array.
- How many times: Once per pop call, no loops or recursion inside pop.
Removing the top item takes the same effort no matter how big the stack is.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | 1 |
| 100 | 1 |
| 1000 | 1 |
Pattern observation: The time stays constant as the stack grows.
Time Complexity: O(1)
This means popping an item takes the same short time no matter how many items are in the stack.
[X] Wrong: "Pop takes longer if the stack is bigger because it has more items to check."
[OK] Correct: Pop only removes the last item directly without checking others, so size does not affect time.
Understanding that pop is a quick, constant-time operation helps you explain stack efficiency clearly in interviews.
"What if the stack was implemented using a linked list instead of an array? How would the time complexity of pop change?"
Practice
pop() operation do in a stack?Solution
Step 1: Understand stack behavior
A stack follows Last In First Out (LIFO) principle, so the last added element is on top.Step 2: Define pop operation
The pop operation removes and returns the top element of the stack, following LIFO.Final Answer:
Removes and returns the top element of the stack -> Option AQuick Check:
pop() = remove top element [OK]
- Thinking pop adds elements
- Confusing pop with peek (which only returns)
- Assuming pop removes bottom element
Solution
Step 1: Recall JavaScript array pop method
JavaScript arrays have a built-in methodpop()to remove the last element.Step 2: Match correct syntax
The correct syntax to pop from an array namedstackisstack.pop();Final Answer:
stack.pop(); -> Option CQuick Check:
Use array.pop() to remove last element [OK]
- Using non-existent methods like popElement()
- Calling pop as a standalone function
- Using remove() which is not valid for arrays
const stack = [10, 20, 30]; const popped = stack.pop(); console.log(popped); console.log(stack);
Solution
Step 1: Understand pop() effect on array
pop() removes the last element (30) and returns it, sopoppedis 30.Step 2: Check stack after pop
After pop, stack becomes [10, 20] because 30 is removed.Final Answer:
30 [10, 20] -> Option DQuick Check:
pop() returns last element and removes it [OK]
- Expecting pop() to return second last element
- Assuming stack remains unchanged after pop
- Confusing pop() return value with stack content
let stack = [1, 2, 3]; let popped = stack.pop; console.log(popped);
Solution
Step 1: Check how pop is used
The code assignsstack.popwithout parentheses, so it assigns the function itself, not the result.Step 2: Correct usage of pop
To pop and get the value, pop must be called with parentheses:stack.pop().Final Answer:
pop method is not called, missing parentheses -> Option BQuick Check:
Use stack.pop() to call pop method [OK]
- Forgetting parentheses when calling pop
- Thinking pop is a property, not a function
- Assuming pop returns undefined without call
stack = [5, 10, 15], what will be the state of the stack after performing two consecutive pop() operations?Solution
Step 1: Perform first pop()
First pop removes 15, stack becomes [5, 10].Step 2: Perform second pop()
Second pop removes 10, stack becomes [5].Final Answer:
[5] -> Option AQuick Check:
Two pops remove last two elements [OK]
- Removing elements from the bottom instead of top
- Assuming pop does not change stack
- Confusing number of pops with elements removed
Ready to prove it?
Pass the test and this pattern moves unknown
