What if you could add items instantly without moving everything else?
Why Push Operation on Stack in DSA Javascript?
Imagine you have a stack of plates at home. You want to add a new plate on top, but you have to carefully lift all plates, move them aside, place the new plate, and then put all plates back. This takes time and effort.
Manually moving each plate every time you add one is slow and easy to mess up. You might drop plates or lose track of the order. It's tiring and error-prone.
The push operation on a stack lets you add a new item on top quickly and safely, without disturbing the rest. It's like having a magic hand that places the plate on top instantly.
let plates = ["plate1", "plate2", "plate3"]; // To add a plate manually: let temp = []; while(plates.length > 0) temp.push(plates.pop()); plates.push("newPlate"); while(temp.length > 0) plates.push(temp.pop());
let stack = []; stack.push("plate1"); stack.push("plate2"); stack.push("newPlate");
It enables fast, safe, and organized addition of items on top, making data handling simple and efficient.
When you use the undo button in a text editor, the actions are stored in a stack. Each new action is pushed on top, so you can undo them in reverse order.
Manual addition to a stack is slow and error-prone.
Push operation adds items quickly on top without disturbing others.
This makes managing last-in-first-out data easy and reliable.
Practice
push operation do in a stack data structure?Solution
Step 1: Understand stack operations
A stack follows Last In First Out (LIFO). Thepushoperation adds an element on top.Step 2: Identify the role of
It inserts a new element at the top, increasing stack size by one.pushFinal Answer:
Adds an element to the top of the stack -> Option DQuick Check:
Push = Add top element [OK]
- Confusing push with pop operation
- Thinking push removes elements
- Mixing stack with queue operations
5 into a stack array named stack in JavaScript?Solution
Step 1: Recall JavaScript array push syntax
JavaScript arrays usepush()method to add elements at the end.Step 2: Match correct method call
stack.push(5);correctly calls push on the array named stack.Final Answer:
stack.push(5); -> Option BQuick Check:
Array push method = stack.push(value) [OK]
- Using non-existent methods like add or insert
- Reversing method and object like push.stack()
- Forgetting parentheses after push
const stack = []; stack.push(10); stack.push(20); stack.push(30); console.log(stack);
Solution
Step 1: Trace push operations on empty stack
Initially stack is empty []. After push(10), stack = [10]. After push(20), stack = [10, 20]. After push(30), stack = [10, 20, 30].Step 2: Understand console.log output
Logging stack prints the array as is: [10, 20, 30].Final Answer:
[10, 20, 30] -> Option CQuick Check:
Push adds at end, so stack = [10,20,30] [OK]
- Assuming push adds at start (top as front)
- Confusing stack order with queue
- Expecting reversed array output
let stack = 5; stack.push(10);
Solution
Step 1: Check variable type of stack
stack is assigned number 5, not an array. Numbers do not have push method.Step 2: Understand push method applicability
push is an array method. Calling push on a number causes error: push is not a function.Final Answer:
stack is not an array, so push method is undefined -> Option AQuick Check:
push only works on arrays [OK]
- Assuming push works on numbers
- Thinking push needs multiple arguments
- Confusing push with add method
[1, 2, 3] onto it so that 3 is on top after all pushes?Choose the correct code snippet.
Solution
Step 1: Understand push order for stack
Push adds elements on top. To have 3 on top, push 1 first, then 2, then 3.Step 2: Analyze options
Sequential pushes of 1 then 2 then 3 add elements individually with 3 on top. Pushing [1,2,3] adds a single array element with [1,2,3] on top. Reverse pushes put 1 on top. Assignment replaces the stack entirely.Final Answer:
stack.push(1); stack.push(2); stack.push(3); -> Option AQuick Check:
Push in order to have last element on top [OK]
- Pushing array [1,2,3] as single element
- Pushing in reverse order
- Replacing stack instead of pushing
Ready to prove it?
Pass the test and this pattern moves unknown
