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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code using XOR to find the unique element?

Given the list arr = [2, 3, 5, 4, 5, 3, 4], what will be printed after running the code below?

DSA Python

arr = [2, 3, 5, 4, 5, 3, 4]
unique = 0
for num in arr:
    unique ^= num
print(unique)

A 0

B 4

C 5

D 2

💡 Hint

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

All numbers except 2 appear twice. XOR cancels out duplicates, leaving 2 as the unique element.

❓ Predict Output

intermediate

What is the output of this XOR code with negative numbers?

Given arr = [-1, -2, -2, -1, -3], what will the code print?

DSA Python

arr = [-1, -2, -2, -1, -3]
unique = 0
for num in arr:
    unique ^= num
print(unique)

A -3

B -1

C 0

D -2

💡 Hint

XOR works the same with negative numbers as with positive.

Pairs cancel out, leaving -3 as the unique number.

🔧 Debug

advanced

What error does this code raise?

What error will this code raise when trying to find the unique element using XOR?

DSA Python

arr = [1, 2, 2, 1, '3']
unique = 0
for num in arr:
    unique ^= num
print(unique)

A No error, prints 3

B SyntaxError

C TypeError

D ValueError

💡 Hint

XOR operator requires both operands to be integers.

The string '3' cannot be XORed with an integer, causing a TypeError.

🧠 Conceptual

advanced

How many times is the XOR operation applied in this code?

Given an array of length n, how many XOR operations does the code below perform?

DSA Python

arr = [some list of length n]
unique = 0
for num in arr:
    unique ^= num
print(unique)

A n

B n-1

C n+1

D 1

💡 Hint

The XOR happens once per element in the loop.

The loop runs n times, applying XOR each time.

🚀 Application

expert

What is the unique element after XOR operations on this large array?

Given the array arr = [10, 14, 10, 14, 20, 30, 20], what is the unique element found by XOR?

DSA Python

arr = [10, 14, 10, 14, 20, 30, 20]
unique = 0
for num in arr:
    unique ^= num
print(unique)

A 20

B 30

C 14

D 10

💡 Hint

Pairs cancel out, leaving the unique number.

All numbers except 30 appear twice, so 30 is the unique element.

Practice

1. What does the XOR operation do when applied between two identical numbers?

easy

A. It returns the number itself

B. It returns 0

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 B

4. Quick Check:

   5 ^ 5 = 0 [OK]

Hint: XOR same numbers always zeroes out [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR
• Assuming XOR sums numbers

2. Which of the following is the correct syntax to find the XOR of all elements in a list arr in Python?

easy

A. result = 0
for num in arr:
result |= num

B. result = 0
for num in arr:
result = result + num

C. result = 0
for num in arr:
result ^= num

D. result = 1
for num in arr:
result *= num

Solution

1. Step 1: Initialize result to 0 for XOR operation

   Start with 0 because XOR with 0 returns the number itself.

2. Step 2: Use ^= operator to XOR each element

   Loop through each number and XOR it with the result to accumulate the XOR of all elements.

3. Final Answer:

   result = 0 for num in arr: result ^= num -> Option C

4. Quick Check:

   Use ^= for XOR accumulation [OK]

Hint: Use ^= operator inside loop for XOR accumulation [OK]

Common Mistakes:

• Using + instead of ^
• Starting result with 1
• Using |= instead of ^=

3. What is the output of the following code?

arr = [2, 3, 5, 4, 5, 3, 4]
result = 0
for num in arr:
    result ^= num
print(result)

medium

A. 2

B. 5

C. 0

D. 4

Solution

1. Step 1: XOR all elements step-by-step

   Start with result=0. XOR with 2 -> 2, XOR with 3 -> 1, XOR with 5 -> 4, XOR with 4 -> 0, XOR with 5 -> 5, XOR with 3 -> 6, XOR with 4 -> 2.

2. Step 2: Final XOR value is the non-repeating element

   The final result after XORing all elements is 2, which appears only once.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Only 2 is non-repeating, XOR result = 2 [OK]

Hint: XOR all elements; result is unique number [OK]

Common Mistakes:

• Not XORing all elements
• Confusing output with repeated numbers
• Assuming sum instead of XOR

4. Identify the error in the following code that tries to find the only non-repeating element using XOR:

def find_unique(arr):
    result = 1
    for num in arr:
        result ^= num
    return result

print(find_unique([1, 2, 2, 1, 3]))

medium

A. The XOR operator should be replaced with +

B. The loop should start from index 1, not 0

C. The function should return sum instead of XOR

D. result should be initialized to 0, not 1

Solution

1. Step 1: Check initialization of result

   Result must start at 0 because XOR with 0 returns the number itself. Starting at 1 changes the result incorrectly.

2. Step 2: Verify loop and XOR usage

   The loop and XOR operator are correct; only initialization is wrong.

3. Final Answer:

   result should be initialized to 0, not 1 -> Option D

4. Quick Check:

   Initialize XOR accumulator with 0 [OK]

Hint: Start XOR accumulator with 0, not 1 [OK]

Common Mistakes:

• Starting result at 1
• Replacing XOR with addition
• Changing loop range incorrectly

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. Initialize result to 0 and XOR it with every element in the array

B. Sort the array and find the element that does not have a duplicate next to it

C. Use a dictionary to count occurrences and find the element with count 1

D. Add all elements and subtract twice the sum of unique elements

Solution

1. Step 1: Use XOR properties for pairs

   XOR of two identical numbers is 0, so pairs cancel out when XORed.

2. Step 2: XOR all elements starting from 0

   Starting with 0, XOR all elements leaves only the unique element because pairs vanish.

3. Step 3: Compare with other methods

   Sorting or dictionary methods work but are less efficient in time or space.

4. Final Answer:

   Initialize result to 0 and XOR it with every element in the array -> Option A

5. Quick Check:

   XOR all elements starting at 0 for unique element [OK]

Hint: XOR all elements starting at 0 to find unique [OK]

Common Mistakes:

• Using sorting which is slower
• Using extra space with dictionary
• Incorrect arithmetic approach