Find the Only Non Repeating Element Using XOR in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust code using XOR to find the unique element?

Consider the Rust code below that finds the only non-repeating element in an array using XOR. What will it print?

DSA Rust

fn find_unique(arr: &[i32]) -> i32 {
    let mut result = 0;
    for &num in arr {
        result ^= num;
    }
    result
}

fn main() {
    let nums = [2, 3, 5, 4, 5, 3, 4];
    println!("{}", find_unique(&nums));
}

A 2

B 5

C 3

D 4

💡 Hint

XOR of a number with itself is zero, and XOR with zero is the number itself.

The XOR operation cancels out all numbers appearing twice, leaving only the unique number. Here, 2 appears once, so the output is 2.

❓ Predict Output

intermediate

What is the output of this Rust code with negative numbers?

What will this Rust program print when finding the unique element using XOR with negative numbers?

DSA Rust

fn find_unique(arr: &[i32]) -> i32 {
    let mut result = 0;
    for &num in arr {
        result ^= num;
    }
    result
}

fn main() {
    let nums = [-1, -2, -2, -3, -1];
    println!("{}", find_unique(&nums));
}

A -1

B -2

C 0

D -3

💡 Hint

XOR works the same with negative numbers as with positive numbers.

Pairs cancel out, leaving -3 as the unique element.

🔧 Debug

advanced

What error does this Rust code produce?

This Rust code tries to find the unique element using XOR but has a mistake. What error will it cause?

DSA Rust

fn find_unique(arr: &[i32]) -> i32 {
    let mut result = 0;
    for num in arr {
        result ^= num;
    }
    result
}

fn main() {
    let nums = [1, 2, 1];
    println!("{}", find_unique(&nums));
}

A Type mismatch error because num is a reference

B No error, prints 2

C Syntax error due to missing semicolon

D Runtime panic due to invalid XOR operation

💡 Hint

Check the type of 'num' in the for loop and how XOR is applied.

The for loop iterates over references (&i32), but XOR expects i32 values. The code must dereference num.

🧠 Conceptual

advanced

Why does XOR help find the unique element in an array where others repeat twice?

Which statement best explains why XOR can find the only non-repeating element in an array where every other element appears exactly twice?

A XOR multiplies all numbers, so the unique number is the product

B XOR of two identical numbers is zero, so pairs cancel out, leaving the unique number

C XOR adds all numbers together, so the unique number is the sum

D XOR sorts the array, so the unique number is at the start

💡 Hint

Think about what happens when you XOR a number with itself.

XOR of a number with itself is zero, so pairs cancel out. XOR with zero returns the number itself, leaving the unique element.

❓ Predict Output

expert

What is the output of this Rust code with multiple unique elements?

What will this Rust program print when the array has two unique elements and others repeated twice?

DSA Rust

fn find_unique(arr: &[i32]) -> i32 {
    let mut result = 0;
    for &num in arr {
        result ^= num;
    }
    result
}

fn main() {
    let nums = [1, 2, 3, 2, 1, 4];
    println!("{}", find_unique(&nums));
}

A 0

B 4

C 7

D 3

💡 Hint

XOR of two unique numbers is their XOR combination, not one of them.

Since there are two unique numbers (3 and 4), XOR returns 3 ^ 4 = 7.

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