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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after updating an array element?

Consider the following TypeScript code that updates an element at a specific index in an array. What will be the printed output?

DSA Typescript

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

A 20 -> 30 -> 99 -> 40 -> null

B 10 -> 20 -> 99 -> 40 -> null

C 10 -> 99 -> 30 -> 40 -> null

D 10 -> 20 -> 30 -> 40 -> null

💡 Hint

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

The array starts as [10, 20, 30, 40]. Updating arr[2] changes the third element from 30 to 99. The printed output shows the updated array elements separated by ' -> ' and ends with ' -> null'.

❓ Predict Output

intermediate

What happens when updating an index out of array bounds?

What will be the output of this TypeScript code snippet?

DSA Typescript

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

A Error: Index out of bounds

B

3
1 -> 2 -> 3 -> null

C

6
1 -> 2 -> 3 -> undefined -> undefined -> 10 -> null

D

6
1 -> 2 -> 3 -> null

💡 Hint

In JavaScript/TypeScript, assigning to an index beyond current length extends the array with empty slots.

Assigning arr[5] = 10 sets the sixth element (index 5) to 10. The array length becomes 6. The missing elements (index 3 and 4) are empty slots (holes), which when joined are treated as empty strings, not 'undefined'. Printing the array shows empty positions for those indices.

🔧 Debug

advanced

Why does this array update not work as expected?

Look at this TypeScript code. It tries to update the element at index 1 but the output is unexpected. What is the problem?

DSA Typescript

const arr = [5, 10, 15];
arr[1] == 20;
console.log(arr.join(' -> ') + ' -> null');

A The update uses '==' instead of '=' so the array is not changed. Output: 5 -> 10 -> 15 -> null

B The index 1 is out of bounds, so no update happens. Output: 5 -> 10 -> 15 -> null

C The array is updated correctly. Output: 5 -> 20 -> 15 -> null

D Syntax error due to wrong operator usage

💡 Hint

Check the operator used for assignment.

The code uses '==' which is a comparison operator, not assignment. So arr[1] is not changed. The output remains the original array.

🧠 Conceptual

advanced

What is the effect of updating an array element by reference?

Given this TypeScript code, what will be the output after updating the nested object inside the array?

DSA Typescript

const arr = [{val: 1}, {val: 2}, {val: 3}];
arr[1].val = 99;
console.log(arr.map(x => x.val).join(' -> ') + ' -> null');

A Error: Cannot assign to property 'val'

B 1 -> 2 -> 3 -> null

C 99 -> 2 -> 3 -> null

D 1 -> 99 -> 3 -> null

💡 Hint

Objects inside arrays are accessed by reference, so updating a property changes the original object.

Updating arr[1].val changes the 'val' property of the second object in the array. The output reflects this updated value.

❓ Predict Output

expert

What is the final state of the array after multiple updates?

Analyze this TypeScript code that updates elements at various indices. What is the final printed output?

DSA Typescript

const arr = [0, 0, 0, 0, 0];
arr[1] = arr[3] = 5;
arr[2] = arr[1] + arr[3];
arr[4] = arr[2] * 2;
console.log(arr.join(' -> ') + ' -> null');

A 0 -> 5 -> 10 -> 5 -> 20 -> null

B 0 -> 5 -> 5 -> 5 -> 10 -> null

C 0 -> 5 -> 10 -> 10 -> 20 -> null

D 0 -> 0 -> 0 -> 0 -> 0 -> null

💡 Hint

Remember that assignments like arr[1] = arr[3] = 5 assign 5 to both indices before using their values.

arr[1] and arr[3] are both set to 5. Then arr[2] = 5 + 5 = 10. Finally, arr[4] = 10 * 2 = 20. The first element remains 0.

Practice

1. What does the expression arr[2] do in an array arr?

easy

A. Accesses the element at index 2 of the array

B. Adds a new element at index 2

C. Removes the element at index 2

D. Returns the length of the array

Solution

1. Step 1: Understand array indexing

   Arrays use zero-based indexing, so index 2 means the third element.

2. Step 2: Access element at index 2

   The expression arr[2] retrieves the element stored at the third position.

3. Final Answer:

   Accesses the element at index 2 of the array -> Option A

4. Quick Check:

   arr[2] means get element at index 2 [OK]

Hint: Remember arrays start at 0, so index 2 is third element [OK]

Common Mistakes:

• Thinking index 2 adds or removes elements
• Confusing index with length
• Assuming indexing starts at 1

2. Which of the following is the correct way to update the element at index 1 in a TypeScript array nums to 10?

easy

A. nums.add(1, 10);

B. nums.update(1, 10);

C. nums.set(1, 10);

D. nums[1] = 10;

Solution

1. Step 1: Recall array element update syntax

   In TypeScript, to update an element at a specific index, use array[index] = value;.

2. Step 2: Apply to given array

   So, nums[1] = 10; correctly updates the element at index 1 to 10.

3. Final Answer:

   nums[1] = 10; -> Option D

4. Quick Check:

   Use bracket notation with assignment for update [OK]

Hint: Use bracket and equals sign to update array element [OK]

Common Mistakes:

• Using non-existent methods like update(), set(), add()
• Confusing update syntax with other languages
• Trying to use push() to update element

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

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

medium

A. [5, 10, 15]

B. [20, 10, 15]

C. [5, 20, 15]

D. [5, 15, 20]

Solution

1. Step 1: Analyze initial array and update

   The array starts as [5, 10, 15]. The statement arr[1] = 20; changes the element at index 1 from 10 to 20.

2. Step 2: Determine final array state

   After update, the array becomes [5, 20, 15]. The console.log prints this updated array.

3. Final Answer:

   [5, 20, 15] -> Option C

4. Quick Check:

   Update index 1 to 20 changes array to [5, 20, 15] [OK]

Hint: Remember index 1 is second element, update it directly [OK]

Common Mistakes:

• Not updating the element, so output stays same
• Mixing up indexes and values
• Assuming update changes order

4. Identify the error in the following TypeScript code that tries to update an array element:

let fruits = ['apple', 'banana', 'cherry'];
fruits(1) = 'orange';

medium

A. Using parentheses instead of square brackets for indexing

B. Array is declared with let instead of const

C. Missing semicolon at the end of the line

D. Trying to assign a string to an array element

Solution

1. Step 1: Check array element update syntax

   Array elements must be accessed or updated using square brackets [], not parentheses ().

2. Step 2: Identify the error in code

   The code uses fruits(1) = 'orange'; which is invalid syntax. It should be fruits[1] = 'orange';.

3. Final Answer:

   Using parentheses instead of square brackets for indexing -> Option A

4. Quick Check:

   Use square brackets [] to access/update array elements [OK]

Hint: Use [] not () to access or update array elements [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Thinking let vs const causes error here
• Ignoring syntax errors due to missing semicolon

5. Given the array let scores = [10, 20, 30, 40, 50];, which code snippet correctly doubles the value at index 3 and then prints the updated array?

hard

A. scores[3] = scores[3] * 2;
console.log(scores);

B. scores[3] *= 2;
console.log(scores);

C. scores.update(3, scores[3] * 2);
console.log(scores);

D. scores[3] = scores[3] + scores[3];
console.log(scores);

Solution

1. Step 1: Understand how to double element at index 3

   Index 3 corresponds to the fourth element (value 40). Doubling means multiplying by 2.

2. Step 2: Check each option for correct syntax and effect

   scores[3] = scores[3] * 2; console.log(scores); uses explicit multiplication and assignment, which works. scores[3] *= 2; console.log(scores); uses shorthand *= operator, also correct. scores.update(3, scores[3] * 2); console.log(scores); uses a non-existent method update(). scores[3] = scores[3] + scores[3]; console.log(scores); adds the element to itself, which also doubles it.

3. Step 3: Identify the best concise and correct option

   scores[3] *= 2; console.log(scores); is the most concise and idiomatic way to double the value at index 3.

4. Final Answer:

   scores[3] *= 2; console.log(scores); -> Option B

5. Quick Check:

   Use *= operator to double value at index 3 [OK]

Hint: Use *= operator to multiply and update in one step [OK]

Common Mistakes:

• Using non-existent update() method
• Forgetting to print the array after update
• Using addition instead of multiplication