Find the Only Non Repeating Element Using XOR in DSA Python

Mental Model

XORing a number with itself cancels it out to zero, so XORing all numbers leaves only the one that appears once.

Analogy: Imagine you have pairs of socks in a drawer. If you take out all socks and pair them up, only the single unmatched sock remains alone.

Array: [2, 3, 2, 4, 3]
XOR process: 0 ⊕ 2 -> 2
          2 ⊕ 3 -> 1
          1 ⊕ 2 -> 3
          3 ⊕ 4 -> 7
          7 ⊕ 3 -> 4
Result: 4 (the only non-repeating element)

Dry Run Walkthrough

Input: array: [2, 3, 2, 4, 3]

Goal: Find the element that appears only once in the array

Step 1: Start with result = 0, XOR with first element 2

result = 0 ⊕ 2 = 2

Why: Initialize result to zero and combine with first number

Step 2: XOR result with second element 3

result = 2 ⊕ 3 = 1

Why: Combine next number to cancel pairs later

Step 3: XOR result with third element 2

result = 1 ⊕ 2 = 3

Why: XOR with repeated number cancels previous 2

Step 4: XOR result with fourth element 4

result = 3 ⊕ 4 = 7

Why: Add new number to result

Step 5: XOR result with fifth element 3

result = 7 ⊕ 3 = 4

Why: XOR with repeated 3 cancels it, leaving only 4

Result:

4 (the only non-repeating element)

Annotated Code

DSA Python

class Solution:
    def find_non_repeating(self, nums):
        result = 0
        for num in nums:
            result ^= num  # XOR accumulates unique element
        return result

if __name__ == "__main__":
    arr = [2, 3, 2, 4, 3]
    sol = Solution()
    print(sol.find_non_repeating(arr))

result ^= num # XOR accumulates unique element

XOR current number with result to cancel duplicates and keep unique

Output

4

Complexity Analysis

Time: O(n) because we traverse the array once

Space: O(1) because we use only one variable for XOR accumulation

vs Alternative: Better than using a hash map which uses O(n) space; XOR uses constant space

Edge Cases

Array with only one element

Returns that single element as it is the only one

DSA Python

No special guard needed; XOR with zero returns the element

Array where all elements except one appear twice

Correctly returns the unique element

DSA Python

XOR operation inherently handles this case

When to Use This Pattern

When you see a problem asking for the single unique element among duplicates, use XOR to cancel pairs efficiently.

Common Mistakes

Mistake: Using addition or subtraction instead of XOR to find unique element
Fix: Use XOR operator because it cancels duplicates correctly

Mistake: Initializing result with first element instead of zero
Fix: Initialize result to zero to correctly accumulate XOR of all elements

Summary

Finds the only element that does not repeat in an array using XOR.

Use when all elements except one appear twice and you want a fast, memory-efficient solution.

XORing duplicates cancels them out, leaving only 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