📖 Scenario: You are managing a list of tasks for your daily routine. You want to add a new urgent task at the beginning of your task list so that it gets done first.
🎯 Goal: Build a program that inserts a new task at the beginning of an existing array of tasks.
📋 What You'll Learn
Create an array called tasks with the exact tasks: 'brush teeth', 'eat breakfast', 'go to work'
Create a variable called newTask with the exact value 'check emails'
Insert newTask at the beginning of the tasks array using the correct array method
Print the tasks array after insertion
💡 Why This Matters
🌍 Real World
Managing task lists or queues where urgent items need to be handled first is common in daily apps and software.
💼 Career
Understanding how to manipulate arrays by adding items at the start is important for front-end and back-end developers handling ordered data.
Progress0 / 4 steps
1
Create the initial array of tasks
Create an array called tasks with these exact strings in order: 'brush teeth', 'eat breakfast', 'go to work'
DSA Javascript
Hint
Use square brackets [] to create an array and separate items with commas.
2
Create the new task variable
Create a variable called newTask and assign it the string 'check emails'
DSA Javascript
Hint
Use const newTask = 'check emails'; to create the variable.
3
Insert the new task at the beginning
Use the unshift method on the tasks array to insert newTask at the beginning
DSA Javascript
Hint
The unshift method adds an item to the start of an array.
4
Print the updated tasks array
Print the tasks array using console.log(tasks) to show the updated list
DSA Javascript
Hint
Use console.log(tasks); to print the array.
Practice
(1/5)
1. What does the unshift() method do in a JavaScript array?
easy
A. Adds one or more elements to the beginning of the array
B. Removes the first element from the array
C. Adds one or more elements to the end of the array
D. Removes the last element from the array
Solution
Step 1: Understand the unshift() method
The unshift() method adds elements to the start of an array, shifting existing elements to the right.
Step 2: Compare with other array methods
push() adds to the end, shift() removes from the start, and pop() removes from the end.
Final Answer:
Adds one or more elements to the beginning of the array -> Option A
Quick Check:
unshift() adds at start = B [OK]
Hint: Remember: unshift = add at front [OK]
Common Mistakes:
Confusing unshift() with push()
Thinking unshift() removes elements
Mixing unshift() with shift()
2. Which of the following is the correct syntax to insert the value 10 at the beginning of array arr?
easy
A. arr.pushFront(10);
B. arr.addFirst(10);
C. arr.unshift(10);
D. arr.insertAt(0, 10);
Solution
Step 1: Recall JavaScript array methods
JavaScript arrays use unshift() to add elements at the start.
Step 2: Check each option
Only arr.unshift(10); is valid JavaScript syntax for this operation.
Final Answer:
arr.unshift(10); -> Option C
Quick Check:
Correct syntax for start insertion = A [OK]
Hint: Use unshift() to add at start, push() for end [OK]
Common Mistakes:
Using non-existent methods like addFirst or pushFront
Confusing unshift() with push()
Trying to use insertAt which is not a JS array method
3. What will be the output of the following code?
let arr = [2, 3, 4];
arr.unshift(1);
console.log(arr);
medium
A. [1, 3, 4]
B. [1, 2, 3, 4]
C. [2, 3, 4, 1]
D. [2, 3, 4]
Solution
Step 1: Understand unshift effect on array
The unshift(1) adds 1 at the start, shifting existing elements right.
Step 2: Trace the array after unshift
Original array: [2, 3, 4] After unshift(1): [1, 2, 3, 4]
Final Answer:
[1, 2, 3, 4] -> Option B
Quick Check:
unshift adds at front = [1, 2, 3, 4] [OK]
Hint: unshift adds element at front, shifts others right [OK]
Common Mistakes:
Thinking unshift adds at end
Ignoring that unshift shifts elements
Expecting original array unchanged
4. Identify the error in the following code that tries to insert 0 at the beginning of array nums:
let nums = [1, 2, 3];
nums.unshift = 0;
console.log(nums);
medium
A. Using assignment instead of calling unshift as a function
B. Array nums is not defined
C. unshift cannot add zero
D. Syntax error due to missing parentheses
Solution
Step 1: Analyze the assignment to unshift property
The code assigns 0 to nums.unshift, overwriting the method instead of calling it.
Step 2: Correct usage of unshift
To add 0 at start, call nums.unshift(0); as a function, not assign a value.
Final Answer:
Using assignment instead of calling unshift as a function -> Option A
Quick Check:
unshift must be called, not assigned = A [OK]
Hint: Call unshift() as function, don't assign it a value [OK]
Common Mistakes:
Assigning value to unshift instead of calling it
Missing parentheses when calling unshift
Thinking unshift is a property, not a method
5. Given the array let data = [5, 6, 7];, which code snippet correctly inserts the numbers 2, 3, 4 at the beginning so that the final array is [2, 3, 4, 5, 6, 7]?
hard
A. data.unshift([2, 3, 4]);
B. data.splice(0, 0, [2, 3, 4]);
C. data.push(2, 3, 4);
D. data.unshift(2, 3, 4);
Solution
Step 1: Understand unshift with multiple arguments
The unshift() method can take multiple arguments and adds them all at the start in order.
Step 2: Analyze each option's effect
data.unshift([2, 3, 4]); adds a single array as first element, resulting in nested array. data.unshift(2, 3, 4); adds 2, 3, 4 individually at start, producing desired array. data.push(2, 3, 4); adds elements at end. data.splice(0, 0, [2, 3, 4]); inserts a single array at index 0, nesting again.
Final Answer:
data.unshift(2, 3, 4); -> Option D
Quick Check:
unshift multiple args adds all at front = D [OK]
Hint: Pass multiple values to unshift() to add all at start [OK]
Common Mistakes:
Passing an array instead of separate values to unshift
Using push() which adds at end
Using splice() incorrectly causing nested arrays
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