Slice and array relationship in Go - Time & Space Complexity

Time Complexity: Slice and array relationship

O(n)

Understanding Time Complexity

When working with slices and arrays in Go, it's important to understand how operations on them affect performance.

We want to know how the time to perform actions grows as the size of the slice or array changes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

func sumSlice(s []int) int {
    total := 0
    for _, v := range s {
        total += v
    }
    return total
}

This code sums all elements in a slice by looping through each item once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each element of the slice.
• How many times: Exactly once per element, so as many times as the slice length.

How Execution Grows With Input

As the slice gets bigger, the number of additions grows directly with its size.

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

Pattern observation: The work grows in a straight line with the number of elements.

Final Time Complexity

Time Complexity: O(n)

This means the time to sum the slice grows directly with the number of elements.

Common Mistake

[X] Wrong: "Accessing elements in a slice is slow because it copies the whole array each time."

[OK] Correct: A slice points to the original array's data, so accessing elements is fast and does not copy the entire array.

Interview Connect

Understanding how slices relate to arrays and how operations scale helps you write efficient Go code and explain your reasoning clearly.

Self-Check

"What if we changed the slice to an array parameter instead? How would the time complexity change?"

Practice

1. In Go, what is the relationship between a slice and the underlying array it references?

easy

A. A slice is a completely independent copy of the array.

B. A slice is a view into an array and shares the same data.

C. A slice stores data separately and does not affect the array.

D. A slice can only reference arrays of fixed size 10.

Solution

1. Step 1: Understand slice basics

   A slice in Go points to an underlying array and does not copy data.

2. Step 2: Relationship between slice and array

   Changes to the slice affect the array and vice versa because they share the same data.

3. Final Answer:

   A slice is a view into an array and shares the same data. -> Option B

4. Quick Check:

   Slice shares array data = B [OK]

Hint: Remember: slices share data with arrays, not copy it [OK]

Common Mistakes:

• Thinking slices copy array data
• Believing slices are independent arrays
• Assuming slices have fixed size

2. Which of the following is the correct way to create a slice from an array arr := [5]int{1,2,3,4,5} that includes elements 2, 3, and 4?

easy

A. slice := arr[1:4]

B. slice := arr[2:5]

C. slice := arr[0:3]

D. slice := arr[3:6]

Solution

1. Step 1: Identify slice indices

   Slice syntax is arr[start:end], including start but excluding end index.

2. Step 2: Select correct indices for elements 2,3,4

   Elements 2,3,4 are at indices 1,2,3, so slice is arr[1:4].

3. Final Answer:

   slice := arr[1:4] -> Option A

4. Quick Check:

   Slice indices 1 to 4 exclude 4 = D [OK]

Hint: Slice includes start index, excludes end index [OK]

Common Mistakes:

• Using wrong start or end index
• Trying to slice beyond array length
• Confusing inclusive/exclusive slice bounds

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

arr := [3]int{10, 20, 30}
slice := arr[:]
slice[1] = 99
fmt.Println(arr)

medium

A. [99 20 30]

B. [10 20 30]

C. [10 99 30]

D. [10 20 99]

Solution

1. Step 1: Understand slice referencing

   slice := arr[:] creates a slice referencing the entire array.

2. Step 2: Modify slice element

   slice[1] = 99 changes the second element of the slice and underlying array.

3. Step 3: Print array

   Printing arr shows the updated array: [10 99 30].

4. Final Answer:

   [10 99 30] -> Option C

5. Quick Check:

   Slice change affects array = A [OK]

Hint: Changing slice changes array too [OK]

Common Mistakes:

• Assuming array stays unchanged
• Confusing slice copy with reference
• Misreading slice syntax

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

arr := [4]int{1, 2, 3, 4}
slice := arr[1:3]
slice[2] = 10
fmt.Println(arr)

medium

A. No error, output is [1 2 10 4]

B. Syntax error in slice declaration

C. Array length mismatch error

D. Index out of range error when assigning to slice[2]

Solution

1. Step 1: Check slice length

   slice := arr[1:3] includes elements at indices 1 and 2, length 2.

2. Step 2: Accessing slice[2]

   slice[2] tries to access third element (index 2), which is out of range.

3. Final Answer:

   Index out of range error when assigning to slice[2] -> Option D

4. Quick Check:

   Slice length 2, index 2 invalid = C [OK]

Hint: Slice length limits valid indices from 0 to len-1 [OK]

Common Mistakes:

• Assuming slice length equals array length
• Using invalid slice indices
• Ignoring runtime panics

5. Given the array arr := [5]int{1, 2, 3, 4, 5}, you create two slices:
slice1 := arr[1:4]
slice2 := arr[2:5]
If you set slice1[1] = 99, what will be the content of slice2?

hard

A. [99 4 5]

B. [3 99 5]

C. [3 4 5]

D. [2 99 5]

Solution

1. Step 1: Identify slice1 and slice2 elements

   slice1 = arr[1:4] = elements at indices 1,2,3 -> [2,3,4]
   slice2 = arr[2:5] = elements at indices 2,3,4 -> [3,4,5]

2. Step 2: Modify slice1[1]

   slice1[1] is element at arr index 2 (value 3), changed to 99.
   So arr becomes [1,2,99,4,5].

3. Step 3: Check slice2 after modification

   slice2 references arr[2], arr[3], arr[4] -> [99,4,5].

4. Final Answer:

   [99 4 5] -> Option A

5. Quick Check:

   Shared array data means slice2 reflects change = B [OK]

Hint: Slices share array; changing one affects overlapping slices [OK]

Common Mistakes:

• Mixing up slice indices and values
• Forgetting slices share underlying array
• Assuming slices are independent copies