Why Two Non Repeating Elements in Array Using XOR in DSA C?

The Big Idea

Discover how a simple XOR trick can instantly spot two unique items hidden among many pairs!

The Scenario

Imagine you have a box full of socks where every sock has a matching pair except for two unique socks. You want to find those two unique socks quickly without checking each pair one by one.

The Problem

Manually checking each sock against all others is slow and tiring. You might miss pairs or check the same socks multiple times, making it easy to make mistakes and waste time.

The Solution

Using XOR, a simple math trick, you can quickly find the two unique socks by combining all socks in a way that pairs cancel out, leaving only the unique ones. This method is fast and error-free.

Before vs After

✗ Before

for (int i = 0; i < n; i++) {
  for (int j = i + 1; j < n; j++) {
    if (arr[i] == arr[j]) {
      // mark as paired
    }
  }
}
// find unpaired elements

✓ After

int xor_all = 0;
for (int i = 0; i < n; i++) {
  xor_all ^= arr[i];
}
// Use xor_all to find two unique elements

What It Enables

This technique lets you find two unique elements in a list instantly, even if the list is very large.

Real Life Example

Finding two unique defective products in a large batch where every other product is identical and paired.

Key Takeaways

Manual pair checking is slow and error-prone.

XOR cancels out pairs, revealing unique elements.

Efficient for large data sets with two unique items.

Practice

1. What is the main purpose of using XOR operation to find two non-repeating elements in an array where every other element repeats twice?

easy

A. To cancel out all repeating elements and isolate the XOR of the two unique elements

B. To sort the array in ascending order

C. To count the frequency of each element

D. To reverse the array elements

Solution

1. Step 1: Understand XOR property for duplicates

   XOR of two identical numbers is 0, so XORing all elements cancels out duplicates.

2. Step 2: Result after XORing all elements

   The result is XOR of the two unique elements because duplicates cancel out.

3. Final Answer:

   To cancel out all repeating elements and isolate the XOR of the two unique elements -> Option A

4. Quick Check:

   XOR cancels duplicates = isolate unique elements [OK]

Hint: XOR duplicates become zero, leaving unique elements XOR [OK]

Common Mistakes:

• Thinking XOR sorts the array
• Using XOR to count frequencies
• Assuming XOR reverses elements

2. Which of the following C code snippets correctly initializes the XOR variable to start finding two non-repeating elements?

easy

A. int xor = 1;

B. int xor = 0;

C. int xor = -1;

D. int xor = NULL;

Solution

1. Step 1: Understand XOR identity element

   XOR with 0 leaves the number unchanged, so start with 0.

2. Step 2: Initialize XOR variable

   Setting xor = 0 allows XORing all elements correctly.

3. Final Answer:

   int xor = 0; -> Option B

4. Quick Check:

   Start XOR with 0 for correct accumulation [OK]

Hint: Initialize XOR with 0 because 0 XOR x = x [OK]

Common Mistakes:

• Initializing XOR with 1 or -1
• Using NULL for integer variable
• Not initializing XOR variable

3. What will be the output of the following C code snippet?

int arr[] = {2, 3, 7, 9, 2, 3};
int xor = 0;
for(int i = 0; i < 6; i++) {
    xor ^= arr[i];
}
printf("%d", xor);

medium

A. 14

B. 9

C. 7

D. 0

Solution

1. Step 1: XOR all elements in the array

   Calculate 2^3=1, 1^7=6, 6^9=15, 15^2=13, 13^3=14.

2. Step 2: Result after XORing all elements

   The final XOR value is 14, which is XOR of the two unique elements 7 and 9.

3. Final Answer:

   14 -> Option A

4. Quick Check:

   XOR all elements = 14 [OK]

Hint: XOR all elements to get XOR of unique pair [OK]

Common Mistakes:

• Stopping XOR early
• Miscomputing XOR values
• Assuming output is one unique element

4. Identify the error in the following code snippet that attempts to find two non-repeating elements using XOR:

int arr[] = {4, 1, 2, 1, 2, 5};
int xor = 0;
for(int i = 0; i < 6; i++) {
    xor ^= arr[i];
}
int set_bit = xor & ~(xor - 1);
int x = 0, y = 0;
for(int i = 0; i < 6; i++) {
    if(arr[i] & set_bit)
        x ^= arr[i];
    else
        y ^= arr[i];
}
printf("%d %d", x, y);

medium

A. Incorrect calculation of set_bit using xor & ~(xor - 1)

B. Loop range should be 0 to 5, not 0 to 6

C. Missing initialization of x and y to zero

D. No error, code works correctly

Solution

1. Step 1: Check set_bit calculation

   set_bit = xor & ~(xor - 1) correctly finds rightmost set bit.

2. Step 2: Verify loops and initialization

   Loops run from 0 to 5 (6 elements), x and y initialized to 0 correctly.

3. Final Answer:

   No error, code works correctly -> Option D

4. Quick Check:

   Code logic and syntax are correct [OK]

Hint: Check bitwise operations and loop ranges carefully [OK]

Common Mistakes:

• Miscomputing set_bit
• Wrong loop boundaries
• Uninitialized variables

5. Given an array where exactly two elements appear once and all others appear twice, how can you find those two unique elements efficiently using XOR? Choose the correct sequence of steps.

hard

A. Sort the array, then scan to find unique elements

B. Use a hash map to count frequencies, then pick elements with count 1

C. XOR all elements, find rightmost set bit in XOR result, divide elements into two groups by that bit, XOR each group separately

D. XOR all elements twice to get unique elements

Solution

1. Step 1: XOR all elements to get XOR of two unique elements

   This cancels out duplicates and leaves XOR of unique pair.

2. Step 2: Find rightmost set bit in XOR result

   This bit differs between the two unique elements.

3. Step 3: Divide elements into two groups based on this bit and XOR each group

   This separates the two unique elements and finds them individually.

4. Final Answer:

   XOR all elements, find rightmost set bit in XOR result, divide elements into two groups by that bit, XOR each group separately -> Option C

5. Quick Check:

   Use XOR and bit partitioning to find unique elements [OK]

Hint: Use XOR and bit mask to split and find unique elements [OK]

Common Mistakes:

• Sorting instead of XOR
• Using hash map (extra space)
• XORing twice without grouping