Array Reversal Techniques in DSA Go - Time & Space Complexity

Time Complexity: Array Reversal Techniques

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse an array changes as the array gets bigger.

How does the number of steps grow when we reverse more elements?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func reverseArray(arr []int) {
    left, right := 0, len(arr)-1
    for left < right {
        arr[left], arr[right] = arr[right], arr[left]
        left++
        right--
    }
}

This code reverses the elements of an array by swapping pairs from the ends moving toward the center.

Identify Repeating Operations

• Primary operation: Swapping two elements in the array.
• How many times: Approximately half the length of the array (n/2 times).

How Execution Grows With Input

As the array size grows, the number of swaps grows roughly half as fast.

Input Size (n)	Approx. Operations (swaps)
10	5
100	50
1000	500

Pattern observation: The operations grow linearly with input size, just half as many swaps as elements.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the array grows directly in proportion to the number of elements.

Common Mistake

[X] Wrong: "Reversing an array takes n squared steps because of swapping."

[OK] Correct: Each swap touches only two elements, and the loop runs about n/2 times, so the steps grow linearly, not quadratically.

Interview Connect

Understanding this helps you explain how simple loops affect performance and shows you can analyze basic array operations clearly.

Self-Check

"What if we used recursion to reverse the array instead? How would the time complexity change?"

Practice

1. What does reversing an array mean in simple terms?

easy

A. Changing the order of elements so the last becomes first and first becomes last

B. Sorting the array in ascending order

C. Removing duplicate elements from the array

D. Adding new elements at the end of the array

Solution

1. Step 1: Understand array order

   An array has elements arranged in a sequence from first to last.

2. Step 2: Define reversal

   Reversing means flipping this order so the last element comes first and the first goes last.

3. Final Answer:

   Changing the order of elements so the last becomes first and first becomes last -> Option A

4. Quick Check:

   Reverse = Flip order [OK]

Hint: Reverse means flip the order of elements [OK]

Common Mistakes:

• Confusing reversal with sorting
• Thinking reversal removes elements
• Believing reversal adds elements

2. Which of the following Go code snippets correctly reverses an array named arr of integers?

easy

A. for i := len(arr)-1; i >= 0; i-- { arr[i] = arr[i] }

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

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

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

Solution

1. Step 1: Identify correct swapping logic

   for i := 0; i < len(arr)/2; i++ { arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i] } swaps elements from start and end moving towards the center, which is the correct reversal method.

2. Step 2: Check other options for errors

   for i := 0; i < len(arr); i++ { arr[i] = arr[len(arr)-1-i] } overwrites elements without swapping, losing data. for i := 0; i < len(arr); i++ { arr[i] = arr[i+1] } shifts elements incorrectly. for i := len(arr)-1; i >= 0; i-- { arr[i] = arr[i] } does nothing effective.

3. Final Answer:

   for i := 0; i < len(arr)/2; i++ { arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i] } -> Option C

4. Quick Check:

   Swap pairs from ends to center = for i := 0; i < len(arr)/2; i++ { arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i] } [OK]

Hint: Swap pairs from start and end until middle [OK]

Common Mistakes:

• Not swapping but overwriting elements
• Looping over full array instead of half
• Ignoring index bounds causing errors

3. What is the output of the following Go code?

arr := []int{1, 2, 3, 4, 5}
for i := 0; i < len(arr)/2; i++ {
    arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i]
}
fmt.Println(arr)

medium

A. [5 4 3 2 1]

B. [1 2 3 4 5]

C. [5 2 3 4 1]

D. [1 5 3 2 4]

Solution

1. Step 1: Trace swaps in loop

   Loop runs i=0 and i=1 (since len=5, half=2). - i=0: swap arr[0] and arr[4] -> arr becomes [5 2 3 4 1] - i=1: swap arr[1] and arr[3] -> arr becomes [5 4 3 2 1]

2. Step 2: Check final array after loop

   After both swaps, array is [5 4 3 2 1]. But the code prints after loop, so output is [5 4 3 2 1].

3. Final Answer:

   [5 4 3 2 1] -> Option A

4. Quick Check:

   Swapped ends inward = reversed array [OK]

Hint: Swap pairs from ends, middle stays same if odd length [OK]

Common Mistakes:

• Only swapping first and last once
• Not swapping pairs correctly
• Confusing output with original array

4. Find the error in this Go code that tries to reverse an array arr:

for i := 0; i <= len(arr)/2; i++ {
    arr[i] = arr[len(arr)-1-i]
}

medium

A. The loop should run only till i < len(arr)/2, not <=

B. It overwrites elements without swapping, losing data

C. The code swaps elements correctly, no error

D. The index len(arr)-1-i is out of range

Solution

1. Step 1: Analyze loop and assignment

   The code assigns arr[i] = arr[len(arr)-1-i] without swapping, so original arr[i] is lost.

2. Step 2: Understand consequence

   Without swapping, the array is partially overwritten and not reversed correctly.

3. Final Answer:

   It overwrites elements without swapping, losing data -> Option B

4. Quick Check:

   Overwrite without swap loses data [OK]

Hint: Always swap pairs, don't overwrite [OK]

Common Mistakes:

• Using <= instead of < in loop
• Not swapping both elements
• Ignoring data loss from overwrite

5. Given a slice arr := []int{1, 2, 3, 4, 5, 6}, write a Go function that reverses the slice in place and returns it. Which of the following is the correct implementation?

hard

A. func reverse(arr []int) []int {
for i := 0; i < len(arr); i++ {
arr[i] = arr[len(arr)-1-i]
}
return arr
}

B. func reverse(arr []int) []int {
for i := 0; i <= len(arr)/2; i++ {
arr[i] = arr[len(arr)-1-i]
}
return arr
}

C. func reverse(arr []int) []int {
for i := len(arr)-1; i >= 0; i-- {
arr[i] = arr[i]
}
return arr
}

D. func reverse(arr []int) []int {
for i := 0; i < len(arr)/2; i++ {
arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i]
}
return arr
}

Solution

1. Step 1: Check correct swapping logic

   func reverse(arr []int) []int { for i := 0; i < len(arr)/2; i++ { arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i] } return arr } swaps elements from start and end moving inward, correctly reversing the slice in place.

2. Step 2: Verify other options

   func reverse(arr []int) []int { for i := 0; i < len(arr); i++ { arr[i] = arr[len(arr)-1-i] } return arr } overwrites without swapping, losing data. func reverse(arr []int) []int { for i := len(arr)-1; i >= 0; i-- { arr[i] = arr[i] } return arr } does nothing effective. func reverse(arr []int) []int { for i := 0; i <= len(arr)/2; i++ { arr[i] = arr[len(arr)-1-i] } return arr } overwrites without swapping and uses <= causing index issues.

3. Final Answer:

   func reverse(arr []int) []int { for i := 0; i < len(arr)/2; i++ { arr[i], arr[len(arr)-1-i] = arr[len(arr)-1-i], arr[i] } return arr } -> Option D

4. Quick Check:

   Swap pairs from ends to center and return [OK]

Hint: Swap pairs up to half length, then return [OK]

Common Mistakes:

• Overwriting elements without swapping
• Looping beyond half length
• Not returning the modified slice