Peek lets you see the top without touching the stack--why struggle when you can just glance?
Why Peek Top Element of Stack in DSA Javascript?
Imagine you have a stack of books on your desk. You want to see the title of the book on the very top without removing it. Doing this by lifting each book one by one is tiring and slow.
Manually checking the top book means moving or removing books, which can cause mistakes like dropping or mixing the order. It wastes time and can damage the stack.
The "peek" operation lets you look at the top book instantly without moving anything. It saves time and keeps the stack safe and organized.
if (stack.length > 0) { let topBook = stack[stack.length - 1]; console.log(topBook); }
function peek(stack) {
return stack[stack.length - 1];
}
console.log(peek(stack));Peek lets you quickly check the top item anytime without changing the stack, making your work faster and safer.
When browsing browser history, peek shows the last visited page without removing it, so you can decide if you want to go back or stay.
Manual checking is slow and risky.
Peek gives instant access to the top element without removal.
It keeps the stack order safe and saves time.
Practice
peek operation do in a stack?Solution
Step 1: Understand stack operations
A stack stores elements with last-in, first-out order. Peek shows the top element without removing it.Step 2: Differentiate peek from pop
Pop removes and returns the top element, peek only returns it without removal.Final Answer:
Returns the top element without removing it -> Option AQuick Check:
Peek = top element without removal [OK]
- Confusing peek with pop operation
- Thinking peek adds elements
- Assuming peek checks emptiness
stack in JavaScript?Solution
Step 1: Understand JavaScript array indexing
JavaScript arrays use zero-based indexing. The last element is at index length - 1.Step 2: Identify correct peek syntax
To peek, access the last element without removing it: stack[stack.length - 1]. stack.pop() removes the element, stack.peek() and stack.top() are not default methods.Final Answer:
stack[stack.length - 1] -> Option CQuick Check:
Peek = stack[stack.length - 1] [OK]
- Using stack.pop() which removes element
- Using non-existent methods like peek() or top()
- Accessing stack[stack.length] which is undefined
const stack = [10, 20, 30]; const top = stack[stack.length - 1]; console.log(top);
Solution
Step 1: Identify the stack elements and length
The stack array is [10, 20, 30]. Its length is 3.Step 2: Calculate the top element index and value
stack.length - 1 = 3 - 1 = 2. stack[2] is 30.Final Answer:
30 -> Option BQuick Check:
Top element = 30 [OK]
- Using stack[stack.length] which is undefined
- Confusing first element with top element
- Printing stack.pop() instead of peek
const stack = [5, 15, 25];
function peek() {
return stack[stack.length];
}
console.log(peek());Solution
Step 1: Analyze the peek function index
stack.length returns the number of elements, but array indices go from 0 to length - 1.Step 2: Correct the index to access top element
Use stack.length - 1 to get the last element. Using stack.length returns undefined.Final Answer:
stack.length should be stack.length - 1 -> Option AQuick Check:
Index = length - 1 for top element [OK]
- Using stack.length as index causing undefined
- Thinking peek removes element
- Missing parentheses in console.log (not in this code)
function safePeek(stack) {
// Your code here
}Solution
Step 1: Check if stack is empty
Before accessing the top element, verify stack length is greater than zero to avoid undefined access.Step 2: Return top element or null safely
If stack is not empty, return stack[stack.length - 1], else return null to indicate empty stack.Final Answer:
return stack.length > 0 ? stack[stack.length - 1] : null; -> Option DQuick Check:
Safe peek uses length check and conditional [OK]
- Accessing stack[stack.length] causing undefined
- Using pop which removes element
- Returning stack[0] which is bottom, not top
Ready to prove it?
Pass the test and this pattern moves unknown
