Array Access and Update at Index in DSA Typescript - Time & Space Complexity

Time Complexity: Array Access and Update at Index

O(1)

Understanding Time Complexity

We want to understand how fast we can get or change a value in an array by its position.

The question is: how does the time to access or update an element change as the array grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

const arr = [10, 20, 30, 40, 50];

// Access element at index 2
const value = arr[2];

// Update element at index 4
arr[4] = 60;
    

This code gets the value at position 2 and changes the value at position 4 in the array.

Identify Repeating Operations

Here, there are no loops or repeated steps.

• Primary operation: Direct access or update at a specific index.
• How many times: Exactly once per access or update.

How Execution Grows With Input

Accessing or updating an element takes the same time no matter how big the array is.

Input Size (n)	Approx. Operations
10	1
100	1
1000	1

Pattern observation: The time stays constant even if the array grows larger.

Final Time Complexity

Time Complexity: O(1)

This means accessing or updating an element by index takes the same small amount of time no matter how big the array is.

Common Mistake

[X] Wrong: "Accessing an element takes longer if the array is bigger because it has to look through all items."

[OK] Correct: Arrays let us jump directly to any position, so the time does not depend on the array size.

Interview Connect

Knowing that array access and update are very fast helps you choose the right data structure and explain your choices clearly in interviews.

Self-Check

"What if we used a linked list instead of an array? How would the time complexity for access and update change?"

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