Array Reversal Techniques in DSA Java - 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.

Specifically, how many steps does it take to reverse all elements?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void reverseArray(int[] arr) {
    int left = 0;
    int right = arr.length - 1;
    while (left < right) {
        int temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}
    

This code reverses the elements of an array by swapping pairs from the ends moving toward 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 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 number of operations grows linearly with input size.

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 n² steps because of swapping elements inside a loop."

[OK] Correct: Each element is swapped only once, so the total steps grow linearly, not quadratically.

Interview Connect

Knowing how to analyze simple loops like array reversal helps build confidence for more complex problems.

Self-Check

"What if we used recursion 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 array reversal meaning

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

2. Step 2: Compare with other options

   Sorting arranges elements by value, removing duplicates deletes repeated items, and adding elements is insertion, not 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 array ends, 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 Java code snippets correctly reverses an integer array arr in-place?

easy

A. for(int i = arr.length - 1; i >= 0; i--) { System.out.print(arr[i]); }

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

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

D. Arrays.sort(arr);

Solution

1. Step 1: Analyze in-place reversal logic

   for(int i = 0; i < arr.length / 2; i++) { int temp = arr[i]; arr[i] = arr[arr.length - 1 - i]; arr[arr.length - 1 - i] = temp; } swaps elements from start and end moving towards center, correctly reversing the array in-place.

2. Step 2: Check other options

   for(int i = 0; i < arr.length; i++) { arr[i] = arr[arr.length - 1 - i]; } overwrites elements incorrectly, for(int i = arr.length - 1; i >= 0; i--) { System.out.print(arr[i]); } only prints reversed array without changing it, Arrays.sort(arr); sorts array, not reverses.

3. Final Answer:

   for(int i = 0; i < arr.length / 2; i++) { int temp = arr[i]; arr[i] = arr[arr.length - 1 - i]; arr[arr.length - 1 - i] = temp; } -> Option C

4. Quick Check:

   Swap pairs from ends = reversal [OK]

Hint: Swap pairs from ends up to middle for reversal [OK]

Common Mistakes:

• Overwriting elements without swapping
• Only printing reversed array without changing it
• Using sort instead of reversal

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

int[] arr = {1, 2, 3, 4, 5};
for(int i = 0; i < arr.length / 2; i++) {
  int temp = arr[i];
  arr[i] = arr[arr.length - 1 - i];
  arr[arr.length - 1 - i] = temp;
}
for(int num : arr) {
  System.out.print(num + " ");
}

medium

A. 5 4 3 2 1

B. 1 2 3 4 5

C. 5 2 3 4 1

D. 1 4 3 2 5

Solution

1. Step 1: Trace the reversal loop

   The loop swaps arr[0] with arr[4], arr[1] with arr[3]. After swaps, array becomes [5,4,3,2,1].

2. Step 2: Print the array elements

   The print loop outputs elements in order: 5 4 3 2 1

3. Final Answer:

   5 4 3 2 1 -> Option A

4. Quick Check:

   Swapped ends = reversed array [OK]

Hint: Swapping ends reverses array; print shows reversed order [OK]

Common Mistakes:

• Not swapping correctly, leaving array unchanged
• Confusing print output with original array
• Off-by-one errors in loop

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

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

medium

A. The temp variable should be declared outside the loop

B. The loop condition should be 'i < arr.length / 2' to avoid swapping middle element twice

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

D. The loop should run from 'i = arr.length - 1' down to 0

Solution

1. Step 1: Analyze loop condition

   The loop uses 'i <= arr.length / 2', which causes the middle element to be swapped twice when array length is odd.

2. Step 2: Correct loop condition

   Changing condition to 'i < arr.length / 2' ensures swaps happen only once per pair, avoiding double swap of middle element.

3. Final Answer:

   The loop condition should be 'i < arr.length / 2' to avoid swapping middle element twice -> Option B

4. Quick Check:

   Loop limit excludes middle for odd length [OK]

Hint: Use '<' not '<=' to avoid double swapping middle element [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Declaring temp outside loop unnecessarily
• Incorrect swap logic swapping wrong elements

5. Given a 2D integer array matrix of size n x n, which code snippet correctly reverses each row of the matrix in-place?

hard

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

B. for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { matrix[i][j] = matrix[i][n - 1 - j]; } }

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

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

Solution

1. Step 1: Understand reversing each row

   Reversing each row means swapping elements in each row from start to end moving towards center.

2. Step 2: Analyze options

   for(int i = 0; i < n; i++) { for(int j = 0; j < n / 2; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[i][n - 1 - j]; matrix[i][n - 1 - j] = temp; } } swaps elements in each row correctly. for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { matrix[i][j] = matrix[i][n - 1 - j]; } } overwrites elements incorrectly. for(int j = 0; j < n; j++) { for(int i = 0; i < n / 2; i++) { int temp = matrix[i][j]; matrix[i][j] = matrix[n - 1 - i][j]; matrix[n - 1 - i][j] = temp; } } swaps columns, not rows. for(int i = 0; i < n; i++) { Arrays.sort(matrix[i]); } sorts rows, not reverses.

3. Final Answer:

   for(int i = 0; i < n; i++) { for(int j = 0; j < n / 2; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[i][n - 1 - j]; matrix[i][n - 1 - j] = temp; } } -> Option D

4. Quick Check:

   Swap pairs in each row for reversal [OK]

Hint: Reverse each row by swapping pairs from ends to center [OK]

Common Mistakes:

• Swapping columns instead of rows
• Overwriting elements without swapping
• Sorting rows instead of reversing