Array Access and Update at Index in DSA Go - 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 at a certain 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.

package main

func accessAndUpdate(arr []int, index int, newValue int) int {
    oldValue := arr[index]    // Access element at index
    arr[index] = newValue     // Update element at index
    return oldValue
}

This code gets the value at a given index in the array and then changes it to a new value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Direct access and update of one element by index.
• How many times: Exactly once per function call, no loops or recursion.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	2 (one access + one update)
100	2 (one access + one update)
1000	2 (one access + one update)

Pattern observation: The number of operations stays the same regardless of array size.

Final Time Complexity

Time Complexity: O(1)

This means accessing or updating an element by index takes constant time, no matter how large the array is.

Common Mistake

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

[OK] Correct: Arrays allow direct access by index, so the computer jumps straight to the position without checking others.

Interview Connect

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

Self-Check

"What if we had to search for a value instead of accessing by index? How would the time complexity change?"

Practice

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

easy

A. Accesses the element at index 3 of the array

B. Accesses the element at index 2 of the array

C. Updates the element at index 2 of the array

D. Deletes the element at index 2 of the array

Solution

1. Step 1: Understand array indexing

   Array indexing starts at 0, so index 2 means the third element.

2. Step 2: Identify the operation

   Using arr[2] accesses (reads) the element at index 2 without changing it.

3. Final Answer:

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

4. Quick Check:

   arr[2] reads element at index 2 [OK]

Hint: Remember arrays start at 0 index [OK]

Common Mistakes:

• Thinking index 2 is the second element
• Confusing access with update
• Assuming indexing starts at 1

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

easy

A. nums[1] = 10

B. nums(1) = 10

C. nums[1] := 10

D. nums.update(1, 10)

Solution

1. Step 1: Recall Go array update syntax

   In Go, to update an element, use array[index] = value.

2. Step 2: Check each option

   nums[1] = 10 uses correct syntax. nums(1) = 10 uses parentheses which is invalid. nums[1] := 10 uses := which is for declaration, not update. nums.update(1, 10) uses a method that does not exist.

3. Final Answer:

   nums[1] = 10 -> Option A

4. Quick Check:

   Use square brackets and = for update [OK]

Hint: Use square brackets and = to update array element [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Using := instead of = for update
• Assuming arrays have update methods

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

package main
import "fmt"
func main() {
  arr := [4]int{5, 10, 15, 20}
  arr[2] = 25
  fmt.Println(arr)
}

medium

A. [5 10 25 20]

B. [5 10 15 20]

C. [25 10 15 20]

D. [5 25 15 20]

Solution

1. Step 1: Understand initial array and update

   Initial array is [5, 10, 15, 20]. The code updates element at index 2 (third element) to 25.

2. Step 2: Apply the update and print

   After update, array becomes [5, 10, 25, 20]. Printing shows this updated array.

3. Final Answer:

   [5 10 25 20] -> Option A

4. Quick Check:

   Index 2 updated to 25 [OK]

Hint: Index 2 means third element update [OK]

Common Mistakes:

• Confusing index 2 with index 1 or 3
• Not updating the array before print
• Expecting commas in printed array

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

package main
func main() {
  arr := [3]int{1, 2, 3}
  arr[3] = 4
}

medium

A. Cannot assign to arr[3] because arrays are immutable

B. Syntax error due to missing import

C. Index out of range error at arr[3]

D. No error, code runs fine

Solution

1. Step 1: Check array size and index used

   Array has size 3 with valid indices 0, 1, 2. Accessing arr[3] is invalid.

2. Step 2: Identify error type

   Accessing arr[3] causes runtime panic: index out of range error.

3. Final Answer:

   Index out of range error at arr[3] -> Option C

4. Quick Check:

   Index must be less than array length [OK]

Hint: Array indices go from 0 to length-1 [OK]

Common Mistakes:

• Assuming arr[3] is valid for size 3 array
• Thinking arrays are immutable in Go
• Ignoring runtime errors

5. Given an array arr := [5]int{1, 2, 3, 4, 5}, which code snippet correctly doubles the value at each index using array access and update?

hard

A. for i := 1; i <= len(arr); i++ { arr[i] = arr[i] * 2 }

B. for i := 0; i < len(arr); i++ { arr[i] *= 2; arr[i+1] = arr[i] }

C. for i := 0; i < len(arr); i++ { arr[i+1] = arr[i] * 2 }

D. for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 }

Solution

1. Step 1: Understand array length and valid indices

   Array length is 5, valid indices are 0 to 4. Loop must run from 0 to 4.

2. Step 2: Analyze each loop snippet

   for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 } loops correctly and updates each element by doubling. for i := 1; i <= len(arr); i++ { arr[i] = arr[i] * 2 } uses i from 1 to 5, arr[5] is invalid index. for i := 0; i < len(arr); i++ { arr[i+1] = arr[i] * 2 } updates arr[i+1], which causes out of range at last iteration. for i := 0; i < len(arr); i++ { arr[i] *= 2; arr[i+1] = arr[i] } updates arr[i+1] inside loop, causing out of range and incorrect logic.

3. Final Answer:

   for i := 0; i < len(arr); i++ { arr[i] = arr[i] * 2 } -> Option D

4. Quick Check:

   Loop from 0 to len-1 and update arr[i] [OK]

Hint: Loop indices 0 to len-1 for safe access [OK]

Common Mistakes:

• Looping beyond array length causing index error
• Updating wrong indices inside loop
• Using <= instead of < in loop condition