Find the Only Non Repeating Element Using XOR in DSA C - Time & Space Complexity

Time Complexity: Find the Only Non Repeating Element Using XOR

O(n)

Understanding Time Complexity

We want to know how the time needed to find the unique element changes as the list grows.

How many steps does it take to find the only number that does not repeat?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int findUnique(int arr[], int n) {
    int result = 0;
    for (int i = 0; i < n; i++) {
        result ^= arr[i];
    }
    return result;
}
    

This code finds the only non-repeating element by XORing all elements in the array.

Identify Repeating Operations

• Primary operation: XOR operation inside a single loop.
• How many times: Exactly once for each element in the array (n times).

How Execution Grows With Input

Each new element adds one XOR operation, so the total steps grow directly with the number of elements.

Input Size (n)	Approx. Operations
10	10 XOR operations
100	100 XOR operations
1000	1000 XOR operations

Pattern observation: The number of operations grows in a straight line as input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the unique element grows directly with the number of elements in the array.

Common Mistake

[X] Wrong: "Since XOR is a simple operation, the time is constant no matter the input size."

[OK] Correct: Even though XOR itself is fast, it must be done for every element, so the total time depends on how many elements there are.

Interview Connect

Understanding this simple but powerful approach shows you can use bitwise operations to solve problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if the array was sorted? Would the time complexity to find the unique element change?"

Practice

1. What does the XOR operation help find in an array where every element except one repeats exactly twice?

easy

A. The smallest element

B. The sum of all elements

C. The largest element

D. The only element that does not repeat

Solution

1. Step 1: Understand XOR properties

   XOR of a number with itself is 0, and XOR with 0 is the number itself.

2. Step 2: Apply XOR on all elements

   XORing all elements cancels out duplicates, leaving only the unique element.

3. Final Answer:

   The only element that does not repeat -> Option D

4. Quick Check:

   XOR cancels duplicates = unique element [OK]

Hint: XOR all elements; duplicates vanish, unique remains [OK]

Common Mistakes:

• Thinking XOR sums elements
• Assuming XOR finds max or min
• Confusing XOR with addition

2. Which of the following is the correct syntax to XOR two integers a and b in C?

easy

A. a | b

B. a & b

C. a ^ b

D. a + b

Solution

1. Step 1: Identify XOR operator in C

   The caret symbol (^) is the XOR operator in C.

2. Step 2: Confirm correct usage

   Using a ^ b performs bitwise XOR between integers a and b.

3. Final Answer:

   a ^ b -> Option C

4. Quick Check:

   ^ is XOR operator in C [OK]

Hint: Use ^ for XOR in C, not & or | [OK]

Common Mistakes:

• Using & instead of ^
• Using | instead of ^
• Using + instead of ^

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

int arr[] = {2, 3, 5, 4, 5, 3, 4};
int result = 0;
for(int i = 0; i < 7; i++) {
    result ^= arr[i];
}
printf("%d", result);

medium

A. 2

B. 5

C. 3

D. 4

Solution

1. Step 1: XOR all elements in the array

   Calculate XOR stepwise: 0^2=2, 2^3=1, 1^5=4, 4^4=0, 0^5=5, 5^3=6, 6^4=2.

2. Step 2: Final XOR result is 2

   After XORing all elements, duplicates cancel out, leaving 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   XOR cancels duplicates = 2 [OK]

Hint: XOR all; duplicates vanish, unique stays [OK]

Common Mistakes:

• Not XORing all elements
• Miscomputing XOR steps
• Assuming sum or max instead

4. Identify the error in this C code snippet that tries to find the unique element using XOR:

int arr[] = {1, 2, 1, 3, 2};
int result = arr[0];
for(int i = 1; i <= 5; i++) {
    result ^= arr[i];
}
printf("%d", result);

medium

A. Loop condition should be < 5, not <= 5

B. Initialize result to 0 instead of arr[0]

C. Use & operator instead of ^

D. Array size is incorrect

Solution

1. Step 1: Check loop bounds

   Array has 5 elements indexed 0 to 4; loop runs i from 1 to 5 inclusive, causing out-of-bounds access at i=5.

2. Step 2: Correct loop condition

   Change loop to i < 5 to avoid accessing arr[5] which is invalid.

3. Final Answer:

   Loop condition should be < 5, not <= 5 -> Option A

4. Quick Check:

   Array index out of bounds fixed by i < 5 [OK]

Hint: Loop till i < array size, not <= [OK]

Common Mistakes:

• Using <= instead of < in loop
• Ignoring array bounds
• Wrong operator used

5. Given an array where every element repeats twice except one element which repeats thrice, can XOR alone find the unique element? If not, what is the correct approach?

hard

A. Yes; XOR works for any repetition count

B. No; use bit counting and modulo 3 to find the unique element

C. No; use sum of elements minus sum of unique elements

D. Yes; XOR twice and then XOR thrice elements

Solution

1. Step 1: Understand XOR limitation

   XOR works only when duplicates appear exactly twice; it fails if duplicates appear thrice.

2. Step 2: Use bit counting method

   Count bits in all numbers, take modulo 3 of each bit count to isolate the unique element's bits.

3. Final Answer:

   No; use bit counting and modulo 3 to find the unique element -> Option B

4. Quick Check:

   XOR fails if duplicates appear thrice; bit count works [OK]

Hint: XOR only works for duplicates twice; use bit count mod 3 otherwise [OK]

Common Mistakes:

• Assuming XOR works for any duplicate count
• Trying to XOR thrice duplicates
• Ignoring bit counting approach