Find the Only Non Repeating Element Using XOR in DSA C++ - Time & Space Complexity

Time Complexity: Find the Only Non Repeating Element Using XOR

O(n)

Understanding Time Complexity

We want to know how fast we can find the single number that appears only once in a list where every other number appears twice.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int findUnique(int arr[], int n) {
    int result = 0;
    for (int i = 0; i < n; i++) {
        result ^= arr[i];
    }
    return result;
}
    

This code uses XOR to find the only element that does not repeat in the array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop that goes through each element once.
• How many times: Exactly n times, where n is the number of elements.

How Execution Grows With Input

As the list gets bigger, the number of steps grows directly with the number of elements.

Input Size (n)	Approx. Operations
10	10 XOR operations
100	100 XOR operations
1000	1000 XOR operations

Pattern observation: The operations increase in a straight line as input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the unique element grows directly with the number of elements in the list.

Common Mistake

[X] Wrong: "Using XOR means the time is constant no matter the list size."

[OK] Correct: Even though XOR is fast, we still need to check every element once, so time grows with the list size.

Interview Connect

Understanding this approach shows you can use clever tricks to solve problems efficiently, a skill valued in real coding challenges.

Self-Check

"What if the array was sorted? Would that change the time complexity of finding the unique element?"

Practice

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

easy

A. It returns 0

B. It returns the number itself

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 A

4. Quick Check:

   X XOR X = 0 [OK]

Hint: XOR of same numbers is always zero [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR or AND
• Assuming XOR returns 1 for identical numbers

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

easy

A. int result = a && b;

B. int result = a ^ b;

C. int result = a | b;

D. int result = a & b;

Solution

1. Step 1: Identify XOR operator in C++

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

2. Step 2: Check syntax correctness

   Using int result = a ^ b; correctly computes XOR of a and b.

3. Final Answer:

   int result = a ^ b; -> Option B

4. Quick Check:

   Use ^ for XOR in C++ [OK]

Hint: Use ^ symbol for XOR in C++ [OK]

Common Mistakes:

• Using && instead of ^ for XOR
• Using | or & which are OR and AND
• Missing semicolon at end

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];
}
std::cout << result << std::endl;

medium

A. 2

B. 0

C. 5

D. 4

Solution

1. Step 1: XOR all elements in the array

   Perform 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: Identify the non-repeating element

   All repeating elements cancel out, leaving 2 as the only non-repeating element.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Only non-repeating element = 2 [OK]

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

Common Mistakes:

• Not XORing all elements correctly
• Confusing output with a repeating element
• Assuming sum instead of XOR

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

int arr[] = {1, 2, 1, 3, 2};
int result;
for(int i = 0; i < 5; i++) {
    result ^= arr[i];
}
std::cout << result << std::endl;

medium

A. XOR operator '^' is used incorrectly; should be '&&'

B. Loop runs one extra iteration causing out-of-bounds access

C. Variable 'result' is not initialized before XOR operation

D. Array size is incorrect causing runtime error

Solution

1. Step 1: Check variable initialization

   The variable 'result' is declared but not initialized, so XOR with garbage value leads to wrong result.

2. Step 2: Correct initialization

   Initialize 'result' to 0 before the loop to correctly accumulate XOR.

3. Final Answer:

   Variable 'result' is not initialized before XOR operation -> Option C

4. Quick Check:

   Initialize XOR accumulator to 0 [OK]

Hint: Always initialize XOR accumulator to 0 [OK]

Common Mistakes:

• Not initializing XOR variable
• Using wrong operator instead of ^
• Incorrect loop bounds

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

B. Sort the array and check neighbors for unique element

C. Use a hash map to count occurrences and find the unique element

D. Initialize a variable to 0 and XOR it with every element in the array once

Solution

1. Step 1: Use XOR properties for unique element

   XORing all elements cancels out duplicates because X XOR X = 0, leaving only the unique element.

2. Step 2: Implement efficient XOR approach

   Initialize a variable to 0 and XOR it with each element in one pass for O(n) time and O(1) space.

3. Final Answer:

   Initialize a variable to 0 and XOR it with every element in the array once -> Option D

4. Quick Check:

   Single pass XOR finds unique element efficiently [OK]

Hint: XOR all elements once; duplicates cancel out [OK]

Common Mistakes:

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