Array Access and Update at Index in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code after updating the array?

Consider the following JavaScript code that updates an array element at a specific index. What will be printed after the update?

DSA Javascript

const arr = [10, 20, 30, 40, 50];
arr[2] = 99;
console.log(arr.join(' -> ') + ' -> null');

A 10 -> 99 -> 30 -> 40 -> 50 -> null

B 10 -> 20 -> 30 -> 40 -> 50 -> null

C 10 -> 20 -> 99 -> 40 -> 50 -> null

D 10 -> 20 -> 99 -> 40 -> null

💡 Hint

Remember that array indices start at 0, so arr[2] refers to the third element.

The code changes the element at index 2 (third element) from 30 to 99. The rest of the array remains unchanged.

❓ Predict Output

intermediate

What is the output after multiple updates on the array?

Look at this JavaScript code that updates multiple elements in an array. What will be printed?

DSA Javascript

const arr = [5, 10, 15, 20, 25];
arr[0] = 50;
arr[4] = 100;
console.log(arr.join(' -> ') + ' -> null');

A 5 -> 10 -> 15 -> 20 -> 25 -> null

B 50 -> 10 -> 15 -> 20 -> 100 -> null

C 50 -> 10 -> 15 -> 20 -> 25 -> null

D 5 -> 10 -> 15 -> 20 -> 100 -> null

💡 Hint

Check which indices are updated and what values they get.

The first element (index 0) is changed from 5 to 50, and the last element (index 4) is changed from 25 to 100.

❓ Predict Output

advanced

What happens when updating an index outside the array length?

Consider this JavaScript code that updates an array at an index beyond its current length. What will be printed?

DSA Javascript

const arr = [1, 2, 3];
arr[5] = 10;
console.log(arr.join(' -> ') + ' -> null');

A 1 -> 2 -> 3 -> undefined -> undefined -> 10 -> null

B 1 -> 2 -> 3 -> 10 -> null

C 1 -> 2 -> 3 -> 0 -> 0 -> 10 -> null

D 1 -> 2 -> 3 -> null

💡 Hint

Think about what happens to the array length and missing elements when assigning beyond current length.

Assigning to index 5 extends the array length to 6. Missing elements (index 3 and 4) become undefined.

🔧 Debug

advanced

What error does this code produce when updating an array?

Look at this JavaScript code snippet. What error will it produce when run?

DSA Javascript

const arr = [1, 2, 3];
arr[1.5] = 10;
console.log(arr.join(' -> ') + ' -> null');

A No error; output: 1 -> 2 -> 3 -> null (property added but not counted as array element)

B No error; output: 1 -> 2 -> 3 -> null

C No error; output: 1 -> 3 -> null

D TypeError: Invalid array index

💡 Hint

Remember that array indices must be integers; what happens if you use a decimal?

Using a non-integer index adds a property to the array object but does not change its length or elements.

🧠 Conceptual

expert

How many elements does the array have after this update?

Given the following JavaScript code, how many elements does the array contain after the update?

DSA Javascript

const arr = [0, 1, 2];
arr[10] = 100;

A Undefined number of elements

B 10 elements

C 3 elements

D 11 elements

💡 Hint

Array length updates to one more than the highest index assigned.

Assigning to index 10 sets array length to 11 (indices 0 to 10). Missing elements are empty slots.

Practice

1. What does arr[2] return if arr = [10, 20, 30, 40]?

easy

A. 40

B. 20

C. 10

D. 30

Solution

1. Step 1: Understand array indexing

   Arrays start at index 0, so index 2 is the third element.

2. Step 2: Identify the element at index 2

   The element at index 2 in [10, 20, 30, 40] is 30.

3. Final Answer:

   30 -> Option D

4. Quick Check:

   Index 2 element = 30 [OK]

Hint: Remember arrays start at zero index [OK]

Common Mistakes:

• Thinking index starts at 1
• Confusing element value with index
• Using negative index without explanation

2. Which of the following is the correct way to update the 3rd element of array nums to 50?

easy

A. nums[3] = 50;

B. nums(3) = 50;

C. nums[2] = 50;

D. nums{2} = 50;

Solution

1. Step 1: Recall array indexing syntax

   In JavaScript, arrays use square brackets with zero-based index.

2. Step 2: Identify correct index for 3rd element

   The 3rd element is at index 2, so nums[2] = 50; is correct.

3. Final Answer:

   nums[2] = 50; -> Option C

4. Quick Check:

   Use square brackets and zero-based index [OK]

Hint: Use square brackets and zero-based index [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using 1-based index
• Using curly braces for arrays

3. What will be the output of the following code?

let arr = [5, 10, 15, 20];
arr[1] = arr[3];
console.log(arr);

medium

A. [5, 20, 15, 20]

B. [5, 10, 15, 20]

C. [20, 10, 15, 20]

D. [5, 15, 15, 20]

Solution

1. Step 1: Understand the assignment operation

   The element at index 1 is set to the value at index 3, which is 20.

2. Step 2: Update the array and print

   Original array: [5, 10, 15, 20]. After update: [5, 20, 15, 20].

3. Final Answer:

   [5, 20, 15, 20] -> Option A

4. Quick Check:

   arr[1] = arr[3] updates second element to 20 [OK]

Hint: Remember assignment copies value at index [OK]

Common Mistakes:

• Confusing indexes
• Expecting arr[3] to change
• Thinking assignment swaps values

4. Find the error in the following code snippet:

let data = [1, 2, 3];
data[3] = 4;
console.log(data);

medium

A. No error, array expands and adds 4 at index 3

B. Array becomes empty after assignment

C. Syntax error due to invalid index

D. Index 3 is out of bounds, causes runtime error

Solution

1. Step 1: Understand JavaScript array behavior on out-of-range index

   Assigning to index 3 adds a new element; arrays auto-expand.

2. Step 2: Check the resulting array

   Original: [1, 2, 3]. After assignment: [1, 2, 3, 4]. No error occurs.

3. Final Answer:

   No error, array expands and adds 4 at index 3 -> Option A

4. Quick Check:

   JS arrays auto-expand on assignment [OK]

Hint: JS arrays grow when assigned out-of-range index [OK]

Common Mistakes:

• Expecting runtime error
• Thinking syntax error occurs
• Assuming array clears on assignment

5. Given let arr = [2, 4, 6, 8, 10]; write code to double the value at index 2 and then swap it with the value at index 4. What is the final array?

hard

A. [2, 4, 10, 8, 6]

B. [2, 4, 20, 8, 12]

C. [2, 4, 20, 8, 6]

D. [2, 4, 10, 8, 12]

Solution

1. Step 1: Double the value at index 2

   Value at index 2 is 6. Doubling it: 6 * 2 = 12. Update arr[2] = 12.

2. Step 2: Swap values at index 2 and index 4

   Value at index 4 is 10. Swap arr[2] and arr[4]: arr[2] = 10, arr[4] = 12.

3. Step 3: Write final array

   Array after operations: [2, 4, 10, 8, 12].

4. Final Answer:

   [2, 4, 10, 8, 12] -> Option B

5. Quick Check:

   Double then swap results in [2,4,10,8,12] [OK]

Hint: Double first, then swap values at given indexes [OK]

Common Mistakes:

• Swapping before doubling
• Mixing up indexes
• Not updating array correctly