Find the Only Non Repeating Element Using XOR in DSA Rust

Mental Model

XORing a number with itself cancels it out, leaving only the unique number.

Analogy: Imagine you have pairs of socks in a drawer. If you remove pairs one by one, only the single unmatched sock remains.

Array: [2, 3, 2, 4, 3]
XOR steps: 0 ⊕ 2 -> 2
2 ⊕ 3 -> 1
1 ⊕ 2 -> 3
3 ⊕ 4 -> 7
7 ⊕ 3 -> 4

Dry Run Walkthrough

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

Goal: Find the single number that does not repeat in the array

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

result = 0 ⊕ 2 = 2

Why: Initialize result to first XOR operation

Step 2: XOR result with second element 3

result = 2 ⊕ 3 = 1

Why: Combine next element to cancel duplicates later

Step 3: XOR result with third element 2

result = 1 ⊕ 2 = 3

Why: XOR with duplicate 2 cancels previous 2

Step 4: XOR result with fourth element 4

result = 3 ⊕ 4 = 7

Why: Add unique element 4 to result

Step 5: XOR result with fifth element 3

result = 7 ⊕ 3 = 4

Why: XOR with duplicate 3 cancels previous 3, leaving 4

Result:

Final result = 4, which is the only non-repeating element

Annotated Code

DSA Rust

fn find_unique(arr: &[i32]) -> i32 {
    let mut result = 0;
    for &num in arr {
        result ^= num; // XOR current number with result to cancel duplicates
    }
    result
}

fn main() {
    let arr = [2, 3, 2, 4, 3];
    let unique = find_unique(&arr);
    println!("{}", unique);
}

result ^= num; // XOR current number with result to cancel duplicates

XOR each element to cancel out duplicates, leaving unique number

Output

4

Complexity Analysis

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

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

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

Edge Cases

Array with only one element

Returns that element as unique

DSA Rust

for &num in arr {

Array where all elements except one appear twice

Correctly returns the single non-repeating element

DSA Rust

result ^= num;

Empty array

Returns 0 as default since no elements to XOR

DSA Rust

let mut result = 0;

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: Starting XOR result with the first element instead of zero
Fix: Initialize result to zero to handle empty arrays and simplify logic

Mistake: Using addition or subtraction instead of XOR
Fix: Use XOR operator because it cancels duplicates perfectly

Summary

Finds the single non-repeating element in an array where all others repeat twice.

Use when you need to find a unique element efficiently without extra space.

XORing duplicates cancels them out, leaving only the unique number.

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 is always 0 because bits cancel out.

2. Step 2: Apply this property to the question

   Since the two numbers are identical, XOR returns 0.

3. Final Answer:

   It returns 0 -> Option A

4. Quick Check:

   X XOR X = 0 [OK]

Hint: XOR of same numbers always zero [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with addition
• Assuming XOR returns 1 for identical numbers

2. Which Rust syntax correctly initializes a variable to zero for XOR operation?

easy

A. let mut result = 0;

B. let result = 0;

C. let mut result = 1;

D. let mut result = None;

Solution

1. Step 1: Identify mutable variable declaration

   We need a mutable variable to update XOR results, so use 'mut'.

2. Step 2: Initialize variable to zero

   Zero is the identity for XOR, so initialize with 0.

3. Final Answer:

   let mut result = 0; -> Option A

4. Quick Check:

   Mutable zero initialization for XOR [OK]

Hint: Use 'let mut var = 0;' for XOR accumulator [OK]

Common Mistakes:

• Not using 'mut' for variable to update
• Initializing with 1 instead of 0
• Using Option type instead of integer

3. What is the output of this Rust code?

fn main() {
    let arr = [2, 3, 5, 4, 5, 3, 4];
    let mut result = 0;
    for &num in &arr {
        result ^= num;
    }
    println!("{}", result);
}

medium

A. 4

B. 5

C. 0

D. 2

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: Identify the non-repeating element

   The final XOR result is 2, which is the only number not repeated.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Final XOR = 2 [OK]

Hint: XOR all numbers; result is unique element [OK]

Common Mistakes:

• Miscomputing XOR steps
• Assuming repeated numbers affect result
• Confusing output with repeated elements

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

fn main() {
    let arr = [1, 2, 1, 3, 2];
    let mut result = 0;
    for num in arr {
        result = result ^ num;
    }
    println!("{}", result);
}

medium

A. Missing '&' in for loop to borrow elements

B. No error, code works correctly

C. Incorrect initialization of result variable

D. Using '=' instead of '^=' operator

Solution

1. Step 1: Check for borrowing in for loop

   Iterating over array by value is allowed; no '&' needed.

2. Step 2: Verify XOR operation and initialization

   Result initialized to 0 and XOR applied correctly with '=' operator.

3. Final Answer:

   No error, code works correctly -> Option B

4. Quick Check:

   Code runs and finds unique element [OK]

Hint: XOR with '=' works; no borrow needed for array iteration [OK]

Common Mistakes:

• Thinking '&' is mandatory in for loop
• Confusing '=' with '^=' operator
• Assuming initialization must be different

5. Given an array where every element repeats twice except one element which repeats thrice, can XOR find the unique element? Why or why not?

hard

A. Yes, XOR works for any number of duplicates

B. No, XOR works only if unique element appears once

C. No, XOR only works if all duplicates appear exactly twice

D. Yes, but only if array length is odd

Solution

1. Step 1: Understand XOR behavior with duplicates

   XOR cancels pairs of identical numbers (duplicates appearing twice).

2. Step 2: Analyze effect of triplicate duplicates

   Triplicates do not cancel out with XOR, so result is incorrect.

3. Final Answer:

   No, XOR only works if all duplicates appear exactly twice -> Option C

4. Quick Check:

   XOR cancels pairs, not triplets [OK]

Hint: XOR cancels pairs only, not triplets [OK]

Common Mistakes:

• Assuming XOR works for any duplicate count
• Confusing unique element count with duplicates
• Ignoring XOR cancellation properties