Find the Only Non Repeating Element Using XOR in DSA C++ - Build from Scratch

Find the Only Non Repeating Element Using XOR

📖 Scenario: Imagine you have a list of numbers where every number appears twice except for one number that appears only once. You want to find that unique number quickly.

🎯 Goal: Build a C++ program that finds the only non-repeating element in a list of integers using the XOR operation.

📋 What You'll Learn

Create a vector called numbers with the exact values: 4, 1, 2, 1, 2

Create an integer variable called unique and initialize it to 0

Use a for loop with variable num to iterate over numbers and update unique by XORing it with num

Print the value of unique to display the only non-repeating element

💡 Why This Matters

🌍 Real World

Finding unique elements quickly is useful in error detection, data cleaning, and network packet analysis.

💼 Career

Understanding bitwise operations like XOR is important for software engineers working with low-level data processing and optimization.

1

Create the list of numbers

Create a vector called numbers with these exact integers: 4, 1, 2, 1, 2

DSA C++

#include 
#include 

int main() {
    // Create vector numbers with 4, 1, 2, 1, 2
    // Your code here

    return 0;
}

Hint

Use std::vector<int> and initialize it with the exact numbers inside curly braces.

2

Initialize the XOR accumulator

Create an integer variable called unique and set it to 0

DSA C++

#include 
#include 

int main() {
    std::vector numbers = {4, 1, 2, 1, 2};
    // Create int variable unique and set to 0
    // Your code here

    return 0;
}

Hint

Use int unique = 0; to start with zero before XOR operations.

3

Apply XOR to find the unique element

Use a for loop with variable num to iterate over numbers and update unique by XORing it with num inside the loop

DSA C++

#include 
#include 

int main() {
    std::vector numbers = {4, 1, 2, 1, 2};
    int unique = 0;

    // Use for loop to XOR unique with each num in numbers
    // Your code here

    return 0;
}

Hint

Use a range-based for loop: for (int num : numbers) and inside it update unique ^= num;.

4

Print the unique element

Write std::cout << unique << std::endl; to print the value of unique

DSA C++

#include 
#include 

int main() {
    std::vector numbers = {4, 1, 2, 1, 2};
    int unique = 0;

    for (int num : numbers) {
        unique ^= num;
    }

    // Print unique
    // Your code here

    return 0;
}

Hint

Use std::cout << unique << std::endl; to display the unique number.

Practice

1. What does the XOR operation do when applied between two identical numbers?

easy

A. It returns 0

B. It returns the number itself

C. It returns 1

D. It returns the sum of the numbers

Solution

1. Step 1: Understand XOR property for identical numbers

   XOR of a number with itself always results in 0 because bits cancel out.

2. Step 2: Apply XOR to example

   For example, 5 XOR 5 = 0.

3. Final Answer:

   It returns 0 -> Option A

4. Quick Check:

   X XOR X = 0 [OK]

Hint: XOR of same numbers is always zero [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR or AND
• Assuming XOR returns 1 for identical numbers

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

easy

A. int result = a && b;

B. int result = a ^ b;

C. int result = a | b;

D. int result = a & b;

Solution

1. Step 1: Identify XOR operator in C++

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

2. Step 2: Check syntax correctness

   Using int result = a ^ b; correctly computes XOR of a and b.

3. Final Answer:

   int result = a ^ b; -> Option B

4. Quick Check:

   Use ^ for XOR in C++ [OK]

Hint: Use ^ symbol for XOR in C++ [OK]

Common Mistakes:

• Using && instead of ^ for XOR
• Using | or & which are OR and AND
• Missing semicolon at end

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];
}
std::cout << result << std::endl;

medium

A. 2

B. 0

C. 5

D. 4

Solution

1. Step 1: XOR all elements in the array

   Perform 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: Identify the non-repeating element

   All repeating elements cancel out, leaving 2 as the only non-repeating element.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Only non-repeating element = 2 [OK]

Hint: XOR all elements; result is unique number [OK]

Common Mistakes:

• Not XORing all elements correctly
• Confusing output with a repeating element
• Assuming sum instead of XOR

4. Identify the error in the following code snippet that tries to find the only non-repeating element using XOR:

int arr[] = {1, 2, 1, 3, 2};
int result;
for(int i = 0; i < 5; i++) {
    result ^= arr[i];
}
std::cout << result << std::endl;

medium

A. XOR operator '^' is used incorrectly; should be '&&'

B. Loop runs one extra iteration causing out-of-bounds access

C. Variable 'result' is not initialized before XOR operation

D. Array size is incorrect causing runtime error

Solution

1. Step 1: Check variable initialization

   The variable 'result' is declared but not initialized, so XOR with garbage value leads to wrong result.

2. Step 2: Correct initialization

   Initialize 'result' to 0 before the loop to correctly accumulate XOR.

3. Final Answer:

   Variable 'result' is not initialized before XOR operation -> Option C

4. Quick Check:

   Initialize XOR accumulator to 0 [OK]

Hint: Always initialize XOR accumulator to 0 [OK]

Common Mistakes:

• Not initializing XOR variable
• Using wrong operator instead of ^
• Incorrect loop bounds

5. Given an array where every element appears twice except one element which appears once, how can you find the unique element efficiently using XOR? Choose the best approach.

hard

A. Add all elements and subtract twice the sum of unique elements

B. Sort the array and check neighbors for unique element

C. Use a hash map to count occurrences and find the unique element

D. Initialize a variable to 0 and XOR it with every element in the array once

Solution

1. Step 1: Use XOR properties for unique element

   XORing all elements cancels out duplicates because X XOR X = 0, leaving only the unique element.

2. Step 2: Implement efficient XOR approach

   Initialize a variable to 0 and XOR it with each element in one pass for O(n) time and O(1) space.

3. Final Answer:

   Initialize a variable to 0 and XOR it with every element in the array once -> Option D

4. Quick Check:

   Single pass XOR finds unique element efficiently [OK]

Hint: XOR all elements once; duplicates cancel out [OK]

Common Mistakes:

• Using sorting which is slower
• Using extra space with hash maps
• Incorrect arithmetic approach