Why Find the Only Non Repeating Element Using XOR in DSA C?

The Big Idea

What if you could find the odd one out in a crowd with just one simple trick?

The Scenario

Imagine you have a box full of socks where every sock has a matching pair except one. You want to find that single sock without checking each pair one by one.

The Problem

Checking each sock manually to find the one without a pair takes a lot of time and can easily lead to mistakes, especially if the box is large. You might lose track or miss a sock.

The Solution

Using XOR, you can quickly find the single sock without a pair by combining all socks in a way that pairs cancel out, leaving only the unique one.

Before vs After

✗ Before

int findUnique(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        int count = 0;
        for (int j = 0; j < size; j++) {
            if (arr[i] == arr[j]) count++;
        }
        if (count == 1) return arr[i];
    }
    return -1;
}

✓ After

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

What It Enables

This method allows you to find the unique element in a list instantly, no matter how large the list is.

Real Life Example

Finding the one defective product in a batch where all others are identical, without checking each product twice.

Key Takeaways

Manual checking is slow and error-prone.

XOR cancels out pairs and reveals the unique element.

Efficient and simple solution for finding a single non-repeating item.

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