Dequeue Operation in DSA Javascript - Time & Space Complexity
We want to understand how long it takes to remove an item from a dequeue (double-ended queue).
Specifically, how the time changes as the number of items grows.
Analyze the time complexity of the dequeue operation removing from the front.
class Dequeue {
constructor() {
this.items = [];
}
dequeueFront() {
if (this.items.length === 0) return null;
return this.items.shift();
}
}
This code removes the first item from the dequeue using the array's shift method.
Look at what happens when we remove from the front.
- Primary operation: Removing the first element with
shift(). - How many times: Each removal shifts all remaining items one position left.
When we remove one item, all other items move one step forward.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 9 moves |
| 100 | About 99 moves |
| 1000 | About 999 moves |
Pattern observation: The number of moves grows roughly the same as the number of items.
Time Complexity: O(n)
This means removing from the front takes longer as the dequeue grows because all items must shift.
[X] Wrong: "Removing from the front is always fast and constant time like popping from the end."
[OK] Correct: Because shifting all elements after removal takes time proportional to the number of items.
Understanding this helps you explain why some dequeue implementations use linked lists to get faster removals from both ends.
"What if we used a linked list instead of an array for the dequeue? How would the time complexity of removing from the front change?"
Practice
dequeue in data structures?Solution
Step 1: Understand the definition of dequeue
A dequeue is a data structure that allows insertion and deletion from both front and rear ends.Step 2: Compare options with definition
A double-ended queue where elements can be added or removed from both ends correctly describes a dequeue. Options A, B, and D describe other structures or incorrect behaviors.Final Answer:
A double-ended queue where elements can be added or removed from both ends -> Option DQuick Check:
Dequeue = Double-ended queue [OK]
- Confusing dequeue with stack or normal queue
- Thinking elements can only be added at one end
- Assuming dequeue is a type of list
5 at the front of a dequeue named dq in JavaScript using an array?Solution
Step 1: Recall JavaScript array methods for front insertion
JavaScript arrays useunshift()to add elements at the front.Step 2: Match method with options
Only dq.unshift(5); usesunshift(5), which is correct. Other options are invalid methods.Final Answer:
dq.unshift(5); -> Option BQuick Check:
Array front insert = unshift() [OK]
- Using pushFront which is not a JS array method
- Confusing push (adds at end) with front insertion
- Assuming addFront or insertFront exist in JS arrays
let dq = []; dq.push(10); dq.unshift(20); dq.push(30); dq.shift(); console.log(dq);
Solution
Step 1: Trace each operation on the dequeue array
Start with empty dq = []. push(10) adds 10 at end: [10]. unshift(20) adds 20 at front: [20, 10]. push(30) adds 30 at end: [20, 10, 30]. shift() removes front element (20): [10, 30].Step 2: Determine final state of dq
After all operations, dq is [10, 30].Final Answer:
[10, 30] -> Option AQuick Check:
After shift(), front removed, dq = [10, 30] [OK]
- Confusing shift() with pop()
- Mixing order of unshift and push
- Forgetting shift() removes front element
let dq = []; dq.push(1); dq.popFront(); console.log(dq);
Solution
Step 1: Check JavaScript array methods used
JavaScript arrays do not have a method namedpopFront(). The correct method to remove front element isshift().Step 2: Identify the error
UsingpopFront()causes a runtime error as it is undefined.Final Answer:
popFront() is not a valid JavaScript array method -> Option AQuick Check:
popFront() invalid in JS arrays [OK]
- Assuming popFront() exists like pop()
- Confusing pop() with front removal
- Not knowing shift() removes front element
dq with elements [5, 10, 15, 20]. You want to remove the element 5 from the front and add 25 at the rear, then remove the element 20 from the rear and add 30 at the front. What will be the final state of dq?Solution
Step 1: Remove front element
Start dq = [5, 10, 15, 20]. Remove front element 5 using shift(): dq = [10, 15, 20]. Rear is now 20.Step 2: Remove rear, add rear, add front
Remove rear element 20 using pop(): dq = [10, 15]. Add 25 at rear using push(): dq = [10, 15, 25]. Add 30 at front using unshift(): dq = [30, 10, 15, 25].Final Answer:
[30, 10, 15, 25] -> Option CQuick Check:
Front removed 5, rear removed 20, added 30 front, 25 rear [OK]
- Removing wrong elements from front or rear
- Mixing order of operations
- Using pop() instead of shift() for front removal
Ready to prove it?
Pass the test and this pattern moves unknown
