Array Insertion at Beginning in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting elements at the beginning?

Consider the following JavaScript code that inserts elements at the beginning of an array. What will be the final state of the array after all insertions?

DSA Javascript

let arr = [3, 4, 5];
arr.unshift(2);
arr.unshift(1);
console.log(arr);

A [2, 1, 3, 4, 5]

B [1, 2, 3, 4, 5]

C [3, 4, 5, 1, 2]

D [5, 4, 3, 2, 1]

💡 Hint

Remember that unshift adds elements to the start of the array in the order they are called.

The unshift method adds elements to the beginning of the array. First, 2 is added before 3,4,5 making [2,3,4,5]. Then 1 is added before 2, making [1,2,3,4,5].

❓ Predict Output

intermediate

What is the output after multiple unshift calls?

What will be printed after running this code?

DSA Javascript

const arr = [];
arr.unshift(10);
arr.unshift(20);
arr.unshift(30);
console.log(arr);

A [30, 20, 10]

B [10, 20, 30]

C [20, 30, 10]

D [10, 30, 20]

💡 Hint

Each unshift adds the new element at the start, pushing existing elements right.

The first unshift adds 10: [10]. Then 20 is added at start: [20,10]. Then 30 at start: [30,20,10].

🔧 Debug

advanced

Why does this code not insert at the beginning correctly?

The code below is intended to insert 0 at the beginning of the array, but it does not work as expected. What is the error?

DSA Javascript

let arr = [1, 2, 3];
arr.push(0);
console.log(arr);

A The variable arr is not declared.

B The array is immutable and cannot be changed.

C The console.log is missing parentheses.

D It uses push which adds to the end, not the beginning.

💡 Hint

Check the difference between push and unshift methods.

push adds elements to the end of the array. To add at the beginning, unshift should be used.

❓ Predict Output

advanced

What is the output after inserting multiple elements at the beginning?

What will be the final array after these insertions?

DSA Javascript

let arr = [4, 5];
arr.unshift(2, 3);
arr.unshift(0, 1);
console.log(arr);

A [4, 5, 0, 1, 2, 3]

B [2, 3, 0, 1, 4, 5]

C [0, 1, 2, 3, 4, 5]

D [0, 1, 4, 5, 2, 3]

💡 Hint

unshift can add multiple elements at once, preserving their order at the start.

First unshift adds 2 and 3 at start: [2,3,4,5]. Then unshift adds 0 and 1 at start: [0,1,2,3,4,5].

❓ Predict Output

expert

What is the output after inserting elements at the beginning inside a loop?

What will be the final array after running this code?

DSA Javascript

let arr = [];
for(let i = 1; i <= 3; i++) {
  arr.unshift(i);
}
console.log(arr);

A [3, 2, 1]

B [1, 2, 3]

C [1, 3, 2]

D [3, 1, 2]

💡 Hint

Each unshift adds the new number at the start, pushing previous elements right.

Loop inserts 1: [1], then 2 at start: [2,1], then 3 at start: [3,2,1].

Practice

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

1. Step 1: Understand the unshift() method

   The unshift() method adds elements to the start of an array, shifting existing elements to the right.

2. Step 2: Compare with other array methods

   push() adds to the end, shift() removes from the start, and pop() removes from the end.

3. Final Answer:

   Adds one or more elements to the beginning of the array -> Option A

4. 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

1. Step 1: Recall JavaScript array methods

   JavaScript arrays use unshift() to add elements at the start.

2. Step 2: Check each option

   Only arr.unshift(10); is valid JavaScript syntax for this operation.

3. Final Answer:

   arr.unshift(10); -> Option C

4. 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

1. Step 1: Understand unshift effect on array

   The unshift(1) adds 1 at the start, shifting existing elements right.

2. Step 2: Trace the array after unshift

   Original array: [2, 3, 4]
   After unshift(1): [1, 2, 3, 4]

3. Final Answer:

   [1, 2, 3, 4] -> Option B

4. 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

1. Step 1: Analyze the assignment to unshift property

   The code assigns 0 to nums.unshift, overwriting the method instead of calling it.

2. Step 2: Correct usage of unshift

   To add 0 at start, call nums.unshift(0); as a function, not assign a value.

3. Final Answer:

   Using assignment instead of calling unshift as a function -> Option A

4. 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

1. Step 1: Understand unshift with multiple arguments

   The unshift() method can take multiple arguments and adds them all at the start in order.

2. 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.

3. Final Answer:

   data.unshift(2, 3, 4); -> Option D

4. 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