Find the Only Non Repeating Element Using XOR in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the XOR operation do when applied between two identical bits?

XOR of two identical bits is 0. For example, 1 XOR 1 = 0 and 0 XOR 0 = 0.

beginner

Why does XOR help find the only non-repeating element in an array where every other element repeats twice?

Because XOR of a number with itself is 0, all repeating elements cancel out, leaving only the non-repeating element.

beginner

What is the initial value of the variable used to store XOR results when finding the non-repeating element?

The initial value is 0 because XOR with 0 returns the other number unchanged.

beginner

Given the array [2, 3, 2, 4, 4], what is the only non-repeating element found using XOR?

The only non-repeating element is 3.

beginner

Write the time complexity of finding the only non-repeating element using XOR in an array of size n.

The time complexity is O(n) because we traverse the array once.

What is the result of XOR between a number and 0?

A The number itself

B 0

C 1

D Undefined

XOR of any number with 0 returns the number itself.

If every element except one appears twice in an array, what does XOR of all elements give?

A Sum of all elements

B The non-repeating element

C 0

D Product of all elements

XOR cancels out pairs of identical elements, leaving the unique element.

What is the initial XOR value used when finding the unique element?

A 0

B The first element of the array

C 1

D Array length

We start with 0 because XOR with 0 does not change the value.

Which of these is NOT true about XOR operation?

A XOR of a number with itself is 0

B XOR is commutative

C XOR of a number with 0 is 0

D XOR is associative

XOR of a number with 0 is the number itself, not 0.

What is the time complexity of the XOR method to find the unique element?

A O(n^2)

B O(log n)

C O(1)

D O(n)

The method requires one pass through the array, so it is O(n).

Explain how XOR helps find the only non-repeating element in an array where every other element repeats twice.

Think about how XOR behaves with pairs of identical numbers.

Describe the steps and time complexity of the algorithm to find the only non-repeating element using XOR.

Focus on the single pass through the array and XOR operation.

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