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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the XOR result variable.

DSA C

int xor_result = [1];

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

A1

B0

C-1

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing xor_result to 1 or -1 causes incorrect XOR results.

Using NULL is invalid for integer initialization.

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

2fill in blank

medium

Complete the code to XOR all elements in the array.

DSA C

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

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

A^

B+

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' adds numbers instead of XORing bits.

Using '&' or '|' performs AND or OR, not XOR.

The XOR operator in C is '^'. It combines bits where only one bit is set.

3fill in blank

hard

Fix the error in finding the rightmost set bit mask.

DSA C

int set_bit_mask = xor_result & [1] xor_result);

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

A~

B!

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise NOT (~) instead of negation.

Using addition or logical NOT causes wrong mask.

To find the rightmost set bit, we AND xor_result with its two's complement (-xor_result).

4fill in blank

hard

Fill both blanks to separate the two unique numbers using the set bit mask.

DSA C

int num1 = 0, num2 = 0;
for (int i = 0; i < n; i++) {
    if (arr[i] & [1]) {
        num1 = num1 [2] arr[i];
    } else {
        num2 = num2 ^ arr[i];
    }
}

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

Aset_bit_mask

B^

C+

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of XOR causes wrong results.

Using '&' in place of XOR accumulates bits incorrectly.

We check if the element has the set bit using 'set_bit_mask'. Then XOR elements in each group separately.

5fill in blank

hard

Fill in the blank to print the two unique numbers in ascending order.

DSA C

if (num1 [1] num2) {
    printf("%d %d\n", num1, num2);
} else {
    printf("%d %d\n", num2, num1);
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' prints numbers in descending order.

Using '==' or '!=' is incorrect for ordering.

We print the smaller number first by comparing num1 and num2 with '<'.

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