Find the Only Non Repeating Element Using XOR in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code snippet?

Given an array where every element repeats twice except one, find the unique element using XOR.

DSA C

int arr[] = {2, 3, 5, 4, 5, 3, 4};
int n = 7;
int unique = 0;
for (int i = 0; i < n; i++) {
    unique ^= arr[i];
}
printf("%d\n", unique);

A 5

B 2

C 3

D 4

💡 Hint

XOR of a number with itself is zero, and XOR with zero is the number itself.

All numbers except 2 appear twice, so XOR cancels them out leaving 2.

❓ Predict Output

intermediate

What is the output of this XOR operation on the array?

Find the unique element in the array using XOR.

DSA C

int arr[] = {10, 14, 10, 14, 15};
int n = 5;
int unique = 0;
for (int i = 0; i < n; i++) {
    unique ^= arr[i];
}
printf("%d\n", unique);

A 15

B 10

C 0

D 14

💡 Hint

XOR cancels pairs, leaving the unique number.

10 and 14 appear twice, so XOR cancels them, leaving 15.

❓ Predict Output

advanced

What is the output of this code with negative numbers?

Find the unique element using XOR in an array with negative and positive integers.

DSA C

int arr[] = {-1, -2, -2, -1, -3};
int n = 5;
int unique = 0;
for (int i = 0; i < n; i++) {
    unique ^= arr[i];
}
printf("%d\n", unique);

A -3

B -1

C -2

D 0

💡 Hint

XOR works with negative numbers as well.

Pairs cancel out, leaving -3 as unique.

🧠 Conceptual

advanced

Why does XOR help find the unique element in an array where all others repeat twice?

Choose the best explanation for why XOR operation can find the only non-repeating element.

A XOR sorts the array and picks the middle element as unique.

B XOR adds all numbers together and divides by two to find the unique number.

C XOR multiplies all numbers and the unique number remains after division.

D XOR of a number with itself is zero, so pairs cancel out, leaving the unique number.

💡 Hint

Think about how XOR behaves with identical numbers.

XOR of two identical numbers is zero, so all pairs cancel out leaving the unique number.

❓ Predict Output

expert

What is the output of this code with a large array?

Find the unique element using XOR in a large array where all elements except one repeat twice.

DSA C

int arr[] = {100, 200, 300, 100, 200, 400, 300, 500, 400};
int n = 9;
int unique = 0;
for (int i = 0; i < n; i++) {
    unique ^= arr[i];
}
printf("%d\n", unique);

A 300

B 400

C 500

D 100

💡 Hint

All numbers except one appear twice.

Pairs cancel out, leaving 500 as unique.

Practice

1. What does the XOR operation help find in an array where every element except one repeats exactly twice?

easy

A. The smallest element

B. The sum of all elements

C. The largest element

D. The only element that does not repeat

Solution

1. Step 1: Understand XOR properties

   XOR of a number with itself is 0, and XOR with 0 is the number itself.

2. Step 2: Apply XOR on all elements

   XORing all elements cancels out duplicates, leaving only the unique element.

3. Final Answer:

   The only element that does not repeat -> Option D

4. Quick Check:

   XOR cancels duplicates = unique element [OK]

Hint: XOR all elements; duplicates vanish, unique remains [OK]

Common Mistakes:

• Thinking XOR sums elements
• Assuming XOR finds max or min
• Confusing XOR with addition

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

easy

A. a | b

B. a & b

C. a ^ b

D. a + b

Solution

1. Step 1: Identify XOR operator in C

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

2. Step 2: Confirm correct usage

   Using a ^ b performs bitwise XOR between integers a and b.

3. Final Answer:

   a ^ b -> Option C

4. Quick Check:

   ^ is XOR operator in C [OK]

Hint: Use ^ for XOR in C, not & or | [OK]

Common Mistakes:

• Using & instead of ^
• Using | instead of ^
• Using + instead of ^

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];
}
printf("%d", result);

medium

A. 2

B. 5

C. 3

D. 4

Solution

1. Step 1: XOR all elements in the array

   Calculate 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: Final XOR result is 2

   After XORing all elements, duplicates cancel out, leaving 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   XOR cancels duplicates = 2 [OK]

Hint: XOR all; duplicates vanish, unique stays [OK]

Common Mistakes:

• Not XORing all elements
• Miscomputing XOR steps
• Assuming sum or max instead

4. Identify the error in this C code snippet that tries to find the unique element using XOR:

int arr[] = {1, 2, 1, 3, 2};
int result = arr[0];
for(int i = 1; i <= 5; i++) {
    result ^= arr[i];
}
printf("%d", result);

medium

A. Loop condition should be < 5, not <= 5

B. Initialize result to 0 instead of arr[0]

C. Use & operator instead of ^

D. Array size is incorrect

Solution

1. Step 1: Check loop bounds

   Array has 5 elements indexed 0 to 4; loop runs i from 1 to 5 inclusive, causing out-of-bounds access at i=5.

2. Step 2: Correct loop condition

   Change loop to i < 5 to avoid accessing arr[5] which is invalid.

3. Final Answer:

   Loop condition should be < 5, not <= 5 -> Option A

4. Quick Check:

   Array index out of bounds fixed by i < 5 [OK]

Hint: Loop till i < array size, not <= [OK]

Common Mistakes:

• Using <= instead of < in loop
• Ignoring array bounds
• Wrong operator used

5. Given an array where every element repeats twice except one element which repeats thrice, can XOR alone find the unique element? If not, what is the correct approach?

hard

A. Yes; XOR works for any repetition count

B. No; use bit counting and modulo 3 to find the unique element

C. No; use sum of elements minus sum of unique elements

D. Yes; XOR twice and then XOR thrice elements

Solution

1. Step 1: Understand XOR limitation

   XOR works only when duplicates appear exactly twice; it fails if duplicates appear thrice.

2. Step 2: Use bit counting method

   Count bits in all numbers, take modulo 3 of each bit count to isolate the unique element's bits.

3. Final Answer:

   No; use bit counting and modulo 3 to find the unique element -> Option B

4. Quick Check:

   XOR fails if duplicates appear thrice; bit count works [OK]

Hint: XOR only works for duplicates twice; use bit count mod 3 otherwise [OK]

Common Mistakes:

• Assuming XOR works for any duplicate count
• Trying to XOR thrice duplicates
• Ignoring bit counting approach