Two Non Repeating Elements in Array Using XOR in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to XOR all elements in the array.

DSA C#

int xorResult = 0;
foreach (int num in arr)
{
    xorResult [1]= num;
}

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

A&

B+

C|

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of XOR.

Using bitwise AND or OR instead of XOR.

XOR operator (^) is used to find the XOR of all elements.

2fill in blank

medium

Complete the code to find the rightmost set bit in xorResult.

DSA C#

int setBit = xorResult & -[1];

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

AxorResult

Bnum

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of xorResult for negative.

Using bitwise OR instead of AND.

Using xorResult & -xorResult isolates the rightmost set bit.

3fill in blank

hard

Fix the error in the loop that separates elements into two groups based on setBit.

DSA C#

int x = 0, y = 0;
foreach (int num in arr)
{
    if ((num & [1]) != 0)
        x ^= num;
    else
        y ^= num;
}

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

AsetBit

BxorResult

Cnum

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using xorResult instead of setBit in condition.

Using 1 which checks only the least significant bit.

Use setBit to check if the bit is set in num.

4fill in blank

hard

Fill both blanks to print the two non-repeating elements.

DSA C#

Console.WriteLine("The two non-repeating elements are: {0} and {1}", [1], [2]);

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

Ax

By

CxorResult

DsetBit

Attempts:

3 left

💡 Hint

Common Mistakes

Printing xorResult or setBit instead of x and y.

x and y hold the two unique elements after separation.

5fill in blank

hard

Fill all three blanks to complete the method that returns the two unique elements as an array.

DSA C#

public static int[] FindTwoUnique(int[] arr)
{
    int xorResult = 0;
    foreach (int num in arr)
        xorResult [1]= num;

    int setBit = xorResult & -[2];

    int x = 0, y = 0;
    foreach (int num in arr)
    {
        if ((num & [3]) != 0)
            x ^= num;
        else
            y ^= num;
    }
    return new int[] { x, y };
}

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

A^

BxorResult

CsetBit

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of XOR.

Using wrong variable for negative.

Using xorResult instead of setBit in condition.

Use XOR (^) to combine elements, xorResult for negative, and setBit for grouping.

Practice

1. What is the main purpose of using XOR operation to find two non-repeating elements in an array where all other elements repeat 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 behavior with 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, isolating their combined difference.

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
• Confusing XOR with counting 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 in an integer array arr?

easy

A. int xor = 0;

B. int xor = arr[0];

C. int xor = -1;

D. int xor = 1;

Solution

1. Step 1: Understand XOR identity element

   XOR with 0 leaves the number unchanged, so start with 0 to accumulate XOR of all elements.

2. Step 2: Initialization correctness

   Initializing xor to 0 ensures correct cumulative XOR of array elements.

3. Final Answer:

   int xor = 0; -> Option A

4. Quick Check:

   Start XOR with 0 to accumulate values correctly [OK]

Hint: Start XOR with 0 because 0 ^ x = x [OK]

Common Mistakes:

• Starting XOR with 1 changes results
• Using negative numbers without reason
• Initializing with first array element causes errors

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

int[] arr = {2, 3, 7, 9, 2, 3};
int xor = 0;
foreach (int num in arr) {
    xor ^= num;
}
Console.WriteLine(xor);

medium

A. 9

B. 0

C. 7

D. 14

Solution

1. Step 1: XOR all elements in the array

   Calculate xor = 2 ^ 3 ^ 7 ^ 9 ^ 2 ^ 3. Since 2 ^ 2 = 0 and 3 ^ 3 = 0, this reduces to 7 ^ 9.

2. Step 2: Calculate XOR of remaining unique elements

   7 in binary: 0111, 9 in binary: 1001 XOR: 0111 ^ 1001 = 1110 (binary) = 14 (decimal).

3. Final Answer:

   14 -> Option D

4. Quick Check:

   Duplicates cancel, XOR unique = 7 ^ 9 = 14 [OK]

Hint: Duplicates XOR to zero, XOR unique elements left [OK]

Common Mistakes:

• Forgetting duplicates cancel out
• Calculating XOR incorrectly
• Assuming output is one unique element

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

int[] arr = {4, 1, 2, 1, 2, 5};
int xor = 0;
foreach (int num in arr) {
    xor ^= num;
}
int setBit = xor & ~(xor - 1);
int x = 0, y = 0;
foreach (int num in arr) {
    if ((num & setBit) == 0)
        x ^= num;
    else
        y ^= num;
}
Console.WriteLine(x + ", " + y);

medium

A. The expression to find setBit is incorrect; it should be 'xor & (-xor)' instead of 'xor & ~(xor - 1)'

B. The initial xor should start with arr[0] instead of 0

C. No error; the code will output the two unique elements correctly

D. The if condition should check '(num & setBit) != 0' instead of '== 0'

Solution

1. Step 1: Check how setBit is calculated

   In two's complement, 'xor & ~(xor - 1)' is equivalent to 'xor & (-xor)', both isolate the rightmost set bit correctly.

2. Step 2: Verify code execution

   arr={4,1,2,1,2,5}, xor=4^5=1, setBit=1. Partitions: x=4 (evens), y=5 (odds). Outputs '4, 5' correctly.

3. Final Answer:

   No error; the code will output the two unique elements correctly -> Option C

4. Quick Check:

   Expressions equivalent, code partitions and XORs correctly [OK]

Hint: xor & ~(xor-1) == xor & -xor for rightmost set bit [OK]

Common Mistakes:

• Thinking 'xor & ~(xor-1)' differs from 'xor & -xor'
• Starting xor with first element instead of 0
• Confusing if condition for partitioning

5. Given an array arr = {10, 20, 10, 30, 40, 20}, write the final two non-repeating elements found using the XOR method. What are these elements?

hard

A. [10, 20]

B. [30, 40]

C. [10, 30]

D. [20, 40]

Solution

1. Step 1: XOR all elements to get combined XOR

   Calculate xor = 10 ^ 20 ^ 10 ^ 30 ^ 40 ^ 20. Duplicates cancel: 10 ^ 10 = 0, 20 ^ 20 = 0. Remaining xor = 30 ^ 40.

2. Step 2: Calculate XOR of 30 and 40

   30 in binary: 00011110, 40 in binary: 00101000 XOR: 00011110 ^ 00101000 = 00110110 (54 decimal). Use this to separate elements and find unique ones: 30 and 40.

3. Final Answer:

   [30, 40] -> Option B

4. Quick Check:

   Duplicates cancel, unique elements are 30 and 40 [OK]

Hint: Duplicates cancel, XOR leaves unique pair [OK]

Common Mistakes:

• Including duplicates as unique
• Mixing up XOR results
• Not cancelling duplicates properly