Array Reversal Techniques in DSA C++ - Time & Space Complexity

Time Complexity: Array Reversal Techniques

O(n)

Understanding Time Complexity

We want to understand how long it takes to reverse an array as the array gets bigger.

How does the work grow when the array size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void reverseArray(int arr[], int n) {
    int start = 0, end = n - 1;
    while (start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Swapping two elements in the array.
• How many times: Approximately n/2 times, where n is the array size.

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, about half the size.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the array grows directly in proportion to the array size.

Common Mistake

[X] Wrong: "Reversing an array takes constant time because we just swap a few elements."

[OK] Correct: We must swap pairs for about half the array length, so the work grows as the array gets bigger.

Interview Connect

Understanding how array reversal scales helps you reason about loops and data manipulation, a key skill in many coding challenges.

Self-Check

"What if we used recursion instead of a loop to reverse the array? 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 the meaning of reversal

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

2. Step 2: Compare with other options

   Sorting arranges elements by value, removing duplicates deletes repeated items, adding elements increases size. None of these change order like reversal.

3. Final Answer:

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

4. Quick Check:

   Reversal = Flip order [OK]

Hint: Reversal flips order, not sorting or removing [OK]

Common Mistakes:

• Confusing reversal with sorting
• Thinking reversal removes elements
• Mixing reversal with adding elements

2. Which of the following C++ code snippets correctly reverses an array arr of size n using a for loop?

easy

A. for(int i = 0; i < n; i++) { arr[i] = arr[i]; }

B. for(int i = 0; i < n; i++) { arr[i] = arr[n - i]; }

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

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

Solution

1. Step 1: Analyze the correct reversal loop

   The loop swaps elements from start and end moving towards the center, stopping at half the array size.

2. Step 2: Check other options for errors

   A does nothing, for(int i = 0; i < n; i++) { arr[i] = arr[n - i]; } accesses out-of-bound index (arr[n - i]), C shifts elements incorrectly.

3. Final Answer:

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

4. Quick Check:

   Swap pairs from ends to middle [OK]

Hint: Swap pairs from ends to middle to reverse [OK]

Common Mistakes:

• Accessing out-of-bound indices
• Not stopping loop at half the array
• Overwriting elements without swapping

3. What is the output of the following C++ code?

int arr[] = {1, 2, 3, 4, 5};
int n = 5;
for(int i = 0; i < n / 2; i++) {
  int temp = arr[i];
  arr[i] = arr[n - i - 1];
  arr[n - i - 1] = temp;
}
for(int i = 0; i < n; i++) {
  std::cout << arr[i] << " ";
}

medium

A. 5 4 3 2 1

B. 1 2 3 4 5

C. 5 2 3 4 1

D. 4 3 2 1 5

Solution

1. Step 1: Trace the swapping loop

   Swaps arr[0] with arr[4], arr[1] with arr[3]. After swaps, arr = {5,4,3,2,1}.

2. Step 2: Print the array elements

   The loop prints all elements in order: 5 4 3 2 1.

3. Final Answer:

   5 4 3 2 1 -> Option A

4. Quick Check:

   Reversed array prints 5 4 3 2 1 [OK]

Hint: Swapping pairs reverses array elements [OK]

Common Mistakes:

• Not swapping correctly
• Loop running too many times
• Printing before reversal

4. Identify the error in the following code that attempts to reverse an array arr of size n:

for(int i = 0; i <= n / 2; i++) {
  int temp = arr[i];
  arr[i] = arr[n - i - 1];
  arr[n - i - 1] = temp;
}

medium

A. No error; the code correctly reverses the array

B. The swap logic is incorrect; it should swap arr[i] with arr[i+1]

C. The loop should run from i = 1 to n / 2 instead of 0

D. The loop condition should be 'i < n / 2' to avoid swapping the middle element twice

Solution

1. Step 1: Analyze the loop condition

   The loop runs while i <= n/2, which includes the middle element when n is odd.

2. Step 2: Understand the effect of swapping middle element twice

   Swapping the middle element with itself is unnecessary and can cause incorrect behavior or redundant operation.

3. Final Answer:

   The loop condition should be 'i < n / 2' to avoid swapping the middle element twice -> Option D

4. Quick Check:

   Loop must stop before middle index [OK]

Hint: Use 'i < n/2' to avoid double swapping middle element [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Swapping wrong elements
• Starting loop from 1 instead of 0

5. Given a 2D array (matrix) of size m x n, which approach correctly reverses each row individually in C++?

hard

A. Swap elements column-wise from first to last column for all rows simultaneously

B. Use a nested loop: for each row, swap elements from start to end using two pointers

C. Transpose the matrix and then reverse the entire matrix as a 1D array

D. Sort each row in descending order to reverse it

Solution

1. Step 1: Understand reversing each row

   Reversing each row means flipping elements in that row only, not affecting other rows.

2. Step 2: Identify correct method

   Using two pointers per row to swap elements from start and end is the correct approach.

3. Step 3: Check incorrect options

   Swapping column-wise mixes rows, transposing changes structure, sorting changes order by value, not position.

4. Final Answer:

   Use a nested loop: for each row, swap elements from start to end using two pointers -> Option B

5. Quick Check:

   Reverse rows by swapping pairs inside each row [OK]

Hint: Reverse each row with two-pointer swaps inside nested loop [OK]

Common Mistakes:

• Trying to reverse columns instead of rows
• Using sorting instead of reversing
• Transposing matrix instead of reversing rows