Find the Only Non Repeating Element Using XOR in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the variable that will hold the XOR result.

DSA C++

int unique = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing unique to 1 or -1 causes wrong results.

We start with 0 because XOR with 0 leaves the number unchanged.

2fill in blank

medium

Complete the for loop to iterate over the array elements.

DSA C++

for (int i = 0; i [1] n; i++) {
    unique ^= arr[i];
}

Drag options to blanks, or click blank then click option'

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= n causes out-of-bounds error.

The loop should run from 0 to n-1, so use < n.

3fill in blank

hard

Fix the error in the XOR operation inside the loop.

DSA C++

unique [1] arr[i];

Drag options to blanks, or click blank then click option'

A-=

B^=

C+=

D*=

Attempts:

3 left

💡 Hint

Common Mistakes

Using += or -= changes the logic and gives wrong answer.

We use XOR assignment (^=) to combine elements.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA C++

int findUnique(int [1][], int [2]) {
    int unique = 0;
    for (int i = 0; i < n; i++) {
        unique ^= arr[i];
    }
    return unique;
}

Drag options to blanks, or click blank then click option'

Aarr

Bn

Csize

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names causes errors.

The function takes an array named arr and its size n.

5fill in blank

hard

Fill all three blanks to complete the main function that calls findUnique and prints the result.

DSA C++

#include <iostream>
using namespace std;

int findUnique(int [1][], int [2]);

int main() {
    int arr[] = {2, 3, 5, 4, 5, 3, 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int result = findUnique([3], n);
    cout << "Unique element is: " << result << endl;
    return 0;
}

Drag options to blanks, or click blank then click option'

Aarr

Barray

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names in function call.

The function prototype and call use 'arr' and 'n' as parameters.

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