Array initialization in Go - Time & Space Complexity

Time Complexity: Array initialization

O(n)

Understanding Time Complexity

When we create an array and fill it with values, the computer does some work for each item.

We want to know how this work grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

func initializeArray(n int) []int {
    arr := make([]int, n)
    for i := 0; i < n; i++ {
        arr[i] = i * 2
    }
    return arr
}

This code creates a slice of size n and fills each position with double its index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for loop that sets each element in the slice.
• How many times: Exactly n times, once for each element.

How Execution Grows With Input

As the slice size grows, the number of steps grows in the same way.

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

Pattern observation: Doubling the slice size doubles the work needed.

Final Time Complexity

Time Complexity: O(n)

This means the time to fill the slice grows directly with its size.

Common Mistake

[X] Wrong: "Initializing an array is instant and does not depend on size."

[OK] Correct: Each element must be set, so bigger arrays take more time.

Interview Connect

Understanding how array initialization scales helps you explain performance clearly in interviews.

Self-Check

"What if we used a built-in function that sets all elements to zero instantly? How would the time complexity change?"

Practice

1. Which of the following correctly declares an array of 5 integers in Go?

easy

A. var arr int[5]

B. var arr [5]int

C. arr := []int{5}

D. arr := [5]int{1, 2, 3, 4}

Solution

1. Step 1: Understand Go array declaration syntax

   In Go, arrays are declared with the syntax: var name [size]type.

2. Step 2: Check each option

   A: var arr int[5] wrong syntax (type before size). B: arr := []int{5} is a slice, not an array. C: var arr [5]int correct syntax. D: arr := [5]int{1, 2, 3, 4} too few initializers (4 < 5), compilation error.

3. Final Answer:

   var arr [5]int -> Option B

4. Quick Check:

   Array declaration = var arr [5]int [OK]

Hint: Remember: var name [size]type declares an array [OK]

Common Mistakes:

• Confusing slices with arrays
• Using wrong syntax like int[5]
• Missing var keyword for array declaration

2. Which of these is the correct way to initialize an array of 3 strings with values "a", "b", "c" in Go?

easy

A. arr := [3]string{"a", "b", "c"}

B. arr := []string{"a", "b", "c"}

C. var arr [3]string = {"a", "b", "c"}

D. arr := [3]string{"a", "b", c}

Solution

1. Step 1: Identify correct array initialization syntax

   Arrays use [size]type{values} syntax. arr := [3]string{"a", "b", "c"} matches this exactly.

2. Step 2: Check other options for errors

   arr := []string{"a", "b", "c"} is a slice, not an array. var arr [3]string = {"a", "b", "c"} uses invalid initialization with = and braces. arr := [3]string{"a", "b", c} misses quotes around "c" causing syntax error.

3. Final Answer:

   arr := [3]string{"a", "b", "c"} -> Option A

4. Quick Check:

   Array init with values = [3]string{"a", "b", "c"} [OK]

Hint: Use [size]type{values} for array initialization [OK]

Common Mistakes:

• Using slice syntax instead of array
• Missing quotes around string elements
• Incorrect use of = with array literals

3. What will be the output of this Go code?

package main
import "fmt"
func main() {
  arr := [4]int{1, 2}
  fmt.Println(arr)
}

medium

A. [1 2 0 0]

B. [1 2]

C. [0 0 0 0]

D. Compilation error

Solution

1. Step 1: Understand partial array initialization

   When fewer elements are provided than the array size, Go fills remaining elements with zero values.

2. Step 2: Apply to given array

   Array size is 4, initialized with 1 and 2. Remaining two elements become 0 by default.

3. Final Answer:

   [1 2 0 0] -> Option A

4. Quick Check:

   Partial init fills rest with zero [OK]

Hint: Uninitialized array elements default to zero [OK]

Common Mistakes:

• Assuming array length matches initializer length
• Expecting error for fewer initial values
• Confusing slices with arrays

4. Identify the error in this Go array initialization:

var arr [3]int = [3]int{1, 2, 3, 4}

medium

A. No error, code is correct

B. Missing var keyword

C. Incorrect type declaration

D. Array size mismatch with initializer

Solution

1. Step 1: Check array size and initializer length

   Array declared with size 3 but initializer has 4 elements.

2. Step 2: Understand Go array rules

   Go requires initializer length to match array size exactly, else compilation error occurs.

3. Final Answer:

   Array size mismatch with initializer -> Option D

4. Quick Check:

   Initializer length must equal array size [OK]

Hint: Initializer length must match array size exactly [OK]

Common Mistakes:

• Ignoring extra elements in initializer
• Assuming Go auto-expands array size
• Confusing slices with arrays

5. You want to create an array of 5 integers where the first two elements are 10 and 20, and the rest are zeros. Which code snippet correctly achieves this and prints the array?

hard

A. arr := [5]int{10, 20, 30}
fmt.Println(arr)

B. arr := [3]int{10, 20}
fmt.Println(arr)

C. arr := [5]int{10, 20, 0, 0, 0}
fmt.Println(arr)

D. arr := []int{10, 20}
fmt.Println(arr)

Solution

1. Step 1: Understand array initialization with explicit zeros

   To ensure the last three elements are zeros explicitly, include them in the initializer.

2. Step 2: Check each option

   B: arr := [3]int{10, 20} fmt.Println(arr) wrong size 3, not 5. A: arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) explicitly sets zeros. C: arr := [5]int{10, 20, 30} fmt.Println(arr) third element 30, not zero. D: arr := []int{10, 20} fmt.Println(arr) is a slice, not array.

3. Step 3: Choose the best explicit solution

   A clearly shows all elements matching the requirement exactly.

4. Final Answer:

   arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) -> Option C

5. Quick Check:

   Explicit zeros in initializer = arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) [OK]

Hint: Explicitly list zeros to avoid confusion [OK]

Common Mistakes:

• Assuming missing elements are always zero without clarity
• Using slices instead of arrays
• Adding unintended values in initializer