Find the Only Non Repeating Element Using XOR in DSA Rust - Build from Scratch

Find the Only Non Repeating Element Using XOR

📖 Scenario: Imagine you have a list of numbers where every number appears twice except one number that appears only once. You want to find that unique number quickly.

🎯 Goal: Build a Rust program that finds the only non-repeating element in a list of integers using the XOR operation.

📋 What You'll Learn

Create a vector called numbers with the exact values: [2, 3, 5, 4, 5, 3, 4]

Create a variable called unique and set it to 0

Use a for loop with variable num to iterate over numbers.iter() and update unique by XORing it with *num

Print the value of unique using println!("{}", unique)

💡 Why This Matters

🌍 Real World

Finding a unique element quickly is useful in error detection, data cleaning, and signal processing.

💼 Career

Understanding XOR and loops is fundamental for software engineers working with algorithms and low-level data manipulation.

1

Create the list of numbers

Create a vector called numbers with the exact values: [2, 3, 5, 4, 5, 3, 4]

DSA Rust

# Create the vector numbers with the given values
# Your code here

Hint

Use let numbers = vec![2, 3, 5, 4, 5, 3, 4]; to create the vector.

2

Initialize the XOR accumulator

Create a variable called unique and set it to 0 inside the main function after the numbers vector.

DSA Rust

fn main() {
    let numbers = vec![2, 3, 5, 4, 5, 3, 4];
    // Initialize unique to 0
    // Your code here
}

Hint

Use let mut unique = 0; to create a mutable variable for XOR accumulation.

3

Apply XOR to find the unique number

Use a for loop with variable num to iterate over numbers.iter() and update unique by XORing it with *num inside the main function.

DSA Rust

fn main() {
    let numbers = vec![2, 3, 5, 4, 5, 3, 4];
    let mut unique = 0;
    // Use a for loop to XOR each number with unique
    // Your code here
}

Hint

Use for num in numbers.iter() and update unique with unique ^= *num;.

4

Print the unique number

Print the value of unique using println!("{}", unique) inside the main function.

DSA Rust

fn main() {
    let numbers = vec![2, 3, 5, 4, 5, 3, 4];
    let mut unique = 0;
    for num in numbers.iter() {
        unique ^= *num;
    }
    // Print the unique number
    // Your code here
}

Hint

Use println!("{}", unique); to display 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