Array Declaration and Initialization in DSA Go - Time & Space Complexity

Time Complexity: Array Declaration and Initialization

O(n)

Understanding Time Complexity

We want to understand how long it takes to create and fill an array when we declare it.

How does the time needed change as the array size grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

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

This code creates a slice of size n and sets each element to zero.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that sets each element to zero.
• How many times: Exactly n times, once for each element.

How Execution Grows With Input

As the array size n grows, the number of steps grows directly with n.

Input Size (n)	Approx. Operations
10	10 steps to set elements
100	100 steps to set elements
1000	1000 steps to set elements

Pattern observation: The work grows in a straight line with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to create and fill the array grows directly with the number of elements.

Common Mistake

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

[OK] Correct: Even though memory is allocated quickly, setting each element takes time proportional to the array size.

Interview Connect

Understanding how array initialization scales helps you reason about performance when working with large data sets in real projects.

Self-Check

"What if we used a built-in function that initializes the array with a default value instead of a loop? How would the time complexity change?"

Practice

1. Which of the following is the correct way to declare an array of 5 integers in Go?

easy

A. var arr []int = make([]int, 5)

B. var arr [5]int

C. arr := [5]int

D. arr := []int{5}

Solution

1. Step 1: Understand array declaration syntax in Go

   Arrays in Go are declared with a fixed size inside square brackets, followed by the type, e.g., var arr [5]int.

2. Step 2: Analyze each option

   var arr [5]int correctly declares an array of 5 integers. arr := [5]int uses shorthand but lacks the array literal {}, causing a syntax error. var arr []int = make([]int, 5) declares a slice, not an array. arr := []int{5} declares a slice with one element 5, not an array of size 5.

3. Final Answer:

   var arr [5]int -> Option B

4. Quick Check:

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

Hint: Arrays use fixed size in brackets, slices do not [OK]

Common Mistakes:

• Confusing slices with arrays
• Using shorthand without initialization
• Incorrect size placement

2. Which of the following is the correct syntax to initialize an array with values 1, 2, 3 in Go?

easy

A. arr := []int{1, 2, 3}

B. arr := [3]int(1, 2, 3)

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

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

Solution

1. Step 1: Review array initialization syntax

   In Go, arrays are initialized with curly braces and values inside, e.g., [3]int{1, 2, 3}.

2. Step 2: Check each option

   var arr = [3]int{1, 2, 3} uses correct syntax. arr := [3]int(1, 2, 3) uses parentheses which is invalid. var arr [3]int = {1, 2, 3} misses the type keyword before braces. arr := []int{1, 2, 3} declares a slice, not an array.

3. Final Answer:

   var arr = [3]int{1, 2, 3} -> Option D

4. Quick Check:

   Array init with braces = var arr = [3]int{1, 2, 3} [OK]

Hint: Use curly braces {} for array initialization [OK]

Common Mistakes:

• Using parentheses instead of braces
• Confusing slices with arrays
• Omitting type in declaration

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

package main
import "fmt"
func main() {
  var arr = [4]int{10, 20}
  fmt.Println(arr)
}

medium

A. [10 20 0 0]

B. [10 20]

C. [0 0 0 0]

D. Compilation error

Solution

1. Step 1: Understand partial array initialization

   When an array is partially initialized, the remaining elements are set to zero values (0 for int).

2. Step 2: Analyze the given array

   The array size is 4, but only first two elements are initialized as 10 and 20. The last two elements default to 0.

3. Final Answer:

   [10 20 0 0] -> Option A

4. Quick Check:

   Partial init fills rest with zeros = [10 20 0 0] [OK]

Hint: Uninitialized array elements default to zero [OK]

Common Mistakes:

• Expecting only initialized elements in output
• Thinking partial init causes error
• Confusing slices with arrays

4. Identify the error in the following Go code snippet:

package main
func main() {
  var arr [3]int = [3]int{1, 2}
}

medium

A. No error, code is correct

B. Array size mismatch error

C. Syntax error in array declaration

D. Missing import statement

Solution

1. Step 1: Check array size and initialization

   The array is declared with size 3 but initialized with 2 values. Go allows this and fills the last element with zero.

2. Step 2: Verify syntax and imports

   No syntax error exists. Import is not mandatory if no external packages are used. The code compiles fine.

3. Final Answer:

   No error, code is correct -> Option A

4. Quick Check:

   Partial init allowed, no import needed [OK]

Hint: Partial initialization is allowed; missing imports only if used [OK]

Common Mistakes:

• Assuming size mismatch causes error
• Expecting import for main package
• Confusing slices with arrays

5. You want to declare an array of 5 integers in Go and initialize only the last element to 100, leaving others as zero. Which code snippet achieves this correctly?

hard

A. var arr = [5]int{0: 100}

B. var arr = [5]int{5: 100}

C. var arr = [5]int{4: 100}

D. var arr = [5]int{100}

Solution

1. Step 1: Understand indexed initialization in arrays

   Go allows initializing specific elements by index using syntax like {index: value}. Indexing starts at 0.

2. Step 2: Analyze each option

   var arr = [5]int{4: 100} initializes the element at index 4 (the last element in a 5-element array) to 100, others default to 0. var arr = [5]int{5: 100} uses index 5 which is out of range. var arr = [5]int{100} initializes only the first element. var arr = [5]int{0: 100} initializes index 0 to 100, not the last.

3. Final Answer:

   var arr = [5]int{4: 100} -> Option C

4. Quick Check:

   Use index 4 for last element in 5-size array [OK]

Hint: Index starts at 0; last element is index size-1 [OK]

Common Mistakes:

• Using out-of-range index
• Initializing first element instead of last
• Omitting index for specific element