Find the Only Non Repeating Element Using XOR in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does XOR operation do when applied between two identical bits?

XOR of two identical bits is 0. For example, 1 XOR 1 = 0 and 0 XOR 0 = 0.

beginner

Why does XOR help find the only non-repeating element in a list where every other element repeats twice?

Because XOR of a number with itself is 0, all repeating elements cancel out, leaving only the non-repeating element after XORing all elements.

beginner

What is the initial value used when XORing all elements to find the non-repeating element?

The initial value is 0 because XOR with 0 leaves the number unchanged.

beginner

Given the list [2, 3, 2, 4, 4], what is the only non-repeating element found using XOR?

The only non-repeating element is 3.

beginner

Write the Python code snippet to find the only non-repeating element using XOR.

def find_non_repeating(arr): result = 0 for num in arr: result ^= num return result

What is the result of XOR between a number and 0?

A 1

B 0

C Undefined

D The number itself

XOR with 0 leaves the number unchanged.

If every element except one repeats twice, what will XOR of all elements give?

A Only the non-repeating element

B Product of all elements

C 0

D Sum of all elements

Repeating elements cancel out, leaving only the non-repeating element.

What is the time complexity of finding the non-repeating element using XOR?

A O(n)

B O(n^2)

C O(log n)

D O(1)

We scan the list once, so time complexity is linear.

Which of these properties of XOR is used in this problem?

A XOR is commutative and associative

B X XOR 0 = X

C All of the above

D X XOR X = 0

All these properties help in cancelling out duplicates and isolating the unique element.

What will be the output of find_non_repeating([5, 1, 5])?

A 5

B 1

C 0

D Error

5 XOR 1 XOR 5 = (5 XOR 5) XOR 1 = 0 XOR 1 = 1

Explain how XOR operation helps find the only non-repeating element in a list where every other element repeats twice.

Think about how pairs of same numbers behave with XOR.

Write a simple Python function to find the only non-repeating element using XOR and explain each step.

Start with 0 and XOR all elements one by one.

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