Two Non Repeating Elements in Array Using XOR in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of XOR-based two unique elements finder

What is the output of the following JavaScript code that finds two non-repeating elements in an array using XOR?

DSA Javascript

function findTwoUnique(arr) {
  let xor = 0;
  for (const num of arr) {
    xor ^= num;
  }
  let setBit = xor & -xor;
  let x = 0, y = 0;
  for (const num of arr) {
    if ((num & setBit) !== 0) {
      x ^= num;
    } else {
      y ^= num;
    }
  }
  return [x, y].sort((a,b) => a-b);
}

console.log(findTwoUnique([2, 3, 7, 9, 2, 3]));

A [7, 9]

B [2, 3]

C [9, 7]

D [3, 2]

💡 Hint

Focus on how XOR cancels out repeating numbers and isolates unique ones.

The code uses XOR to cancel out all repeating numbers. The remaining XOR value has bits set where the two unique numbers differ. Using the rightmost set bit, it separates numbers into two groups and XORs each group to find the unique numbers. The final result is sorted ascending.

🧠 Conceptual

intermediate

Why XOR helps find two unique elements?

Why does XOR operation help in finding two non-repeating elements in an array where all other elements repeat twice?

A XOR of two identical numbers is zero, so repeating elements cancel out, leaving XOR of unique elements.

B XOR adds all numbers together to find the sum of unique elements.

C XOR multiplies all numbers to isolate unique elements.

D XOR sorts the array to separate unique elements.

💡 Hint

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

XOR of a number with itself is zero, so pairs cancel out. The XOR of the entire array leaves XOR of the two unique numbers, which helps to separate them.

🔧 Debug

advanced

Identify the error in XOR unique elements code

What error will this code produce when trying to find two unique elements using XOR?

DSA Javascript

function findUnique(arr) {
  let xor = 0;
  for (let i = 0; i <= arr.length; i++) {
    xor ^= arr[i];
  }
  return xor;
}

console.log(findUnique([1, 2, 1, 3, 2, 5]));

A ReferenceError: xor is not defined

B SyntaxError: Unexpected token

C TypeError: Cannot read property 'undefined' of undefined

D Output: 6

💡 Hint

Check the loop boundary condition carefully.

No error is produced. The loop runs from i = 0 to i = arr.length (6) inclusive, so arr[6] is undefined. In JavaScript, array access beyond length returns undefined, and undefined is coerced to 0 for bitwise XOR (ToInt32(undefined) = 0). The XOR of elements is 1 ^ 2 ^ 1 ^ 3 ^ 2 ^ 5 = 6, and 6 ^ 0 = 6.

🚀 Application

advanced

Number of items in result after XOR partition

Given the array [4, 1, 2, 1, 2, 5], after finding the rightmost set bit of XOR of all elements and partitioning the array into two groups based on this bit, how many elements will be in the group where the bit is set?

A 2

B 5

C 3

D 4

💡 Hint

Calculate XOR of all elements first, then find the rightmost set bit and count elements with that bit set.

XOR of all elements is 4 ^ 1 ^ 2 ^ 1 ^ 2 ^ 5 = 4 ^ 5 = 1 (binary 0001). The rightmost set bit is bit 0. Elements with bit 0 set are 1, 1, 5, so count is 3.

❓ Predict Output

expert

Output of XOR two unique elements with negative numbers

What is the output of this JavaScript code that finds two unique elements in an array containing negative numbers using XOR?

DSA Javascript

function findTwoUnique(arr) {
  let xor = 0;
  for (const num of arr) {
    xor ^= num;
  }
  let setBit = xor & -xor;
  let x = 0, y = 0;
  for (const num of arr) {
    if ((num & setBit) !== 0) {
      x ^= num;
    } else {
      y ^= num;
    }
  }
  return [x, y].sort((a,b) => a-b);
}

console.log(findTwoUnique([-1, -2, -1, -3, -2, -4]));

A [-2, -1]

B [-4, -3]

C [-1, -2]

D [-3, -4]

💡 Hint

XOR works the same with negative numbers; focus on bitwise operations.

The code correctly finds the two unique negative numbers -4 and -3. The result is sorted ascending as [-4, -3].

Practice

1. What is the main purpose of using XOR operation to find two non-repeating elements in an array?

easy

A. To cancel out all repeating elements and isolate the unique ones

B. To sort the array in ascending order

C. To count the total number of elements in the array

D. To reverse the array elements

Solution

1. Step 1: Understand XOR behavior with duplicates

   XOR of two identical numbers is 0, so repeating elements cancel out.

2. Step 2: Isolate unique elements using XOR

   XOR of all elements leaves XOR of two unique elements, isolating them.

3. Final Answer:

   To cancel out all repeating elements and isolate the unique ones -> Option A

4. Quick Check:

   XOR cancels duplicates = isolate unique elements [OK]

Hint: XOR cancels pairs, leaving only unique elements [OK]

Common Mistakes:

• Thinking XOR sorts the array
• Using XOR to count elements
• Assuming XOR reverses arrays

2. Which of the following JavaScript code snippets correctly initializes the XOR variable to find two non-repeating elements?

easy

A. let xor = null;

B. let xor = 1;

C. let xor = 0;

D. let xor = undefined;

Solution

1. Step 1: Understand XOR identity element

   XOR with 0 returns the other number unchanged, so start with 0.

2. Step 2: Initialize XOR variable correctly

   Initialize xor as 0 to accumulate XOR of all elements properly.

3. Final Answer:

   let xor = 0; -> Option C

4. Quick Check:

   Start XOR with 0 for correct accumulation [OK]

Hint: Initialize XOR with 0, not 1 or null [OK]

Common Mistakes:

• Starting XOR with 1 changes results
• Using null or undefined causes errors
• Not initializing XOR variable

3. What will be the output of the following JavaScript code?

const arr = [2, 3, 7, 9, 2, 3];
let xor = 0;
for (const num of arr) {
  xor ^= num;
}
console.log(xor);

medium

A. 7

B. 0

C. 9

D. 14

Solution

1. Step 1: XOR all elements in the array

   Calculate XOR stepwise: 0^2=2, 2^3=1, 1^7=6, 6^9=15, 15^2=13, 13^3=14.

2. Step 2: Final XOR value

   The final XOR value after processing all elements is 14.

3. Final Answer:

   14 -> Option D

4. Quick Check:

   XOR all elements = 14 [OK]

Hint: XOR duplicates cancel; XOR unique pairs remain [OK]

Common Mistakes:

• Miscomputing XOR stepwise
• Assuming output is one unique element
• Confusing XOR with sum

4. Identify the error in the following code snippet that attempts to find two non-repeating elements using XOR:

const arr = [4, 1, 2, 1, 2, 5];
let xor = 0;
for (let i = 0; i <= arr.length; i++) {
  xor ^= arr[i];
}
console.log(xor);

medium

A. Use let xor = null instead of 0

B. Loop condition should be i < arr.length, not i <= arr.length

C. Array elements should be sorted before XOR

D. XOR should start from 1, not 0

Solution

1. Step 1: Check loop boundary condition

   The loop runs from i=0 to i<=arr.length, which accesses arr[arr.length] (undefined).

2. Step 2: Correct loop condition

   Change condition to i < arr.length to avoid accessing undefined and runtime error.

3. Final Answer:

   Loop condition should be i < arr.length, not i <= arr.length -> Option B

4. Quick Check:

   Loop must stay within array bounds [OK]

Hint: Use i < arr.length to avoid out-of-bounds errors [OK]

Common Mistakes:

• Using <= instead of < in loop
• Starting XOR with 1 instead of 0
• Sorting array unnecessarily

5. Given the array [6, 1, 3, 6, 1, 5], which two numbers are the non-repeating elements found using the XOR method? Choose the correct pair.

hard

A. [3, 5]

B. [6, 1]

C. [1, 3]

D. [5, 6]

Solution

1. Step 1: XOR all elements to get xor of two unique numbers

   6^1=7, 7^3=4, 4^6=2, 2^1=3, 3^5=6 (xor = 6).

2. Step 2: Find rightmost set bit and separate elements

   Rightmost set bit in 6 (binary 110) is bit 2 (value 2). Separate elements into two groups based on this bit and XOR each group to find unique numbers 3 and 5.

3. Final Answer:

   [3, 5] -> Option A

4. Quick Check:

   Unique elements are 3 and 5 [OK]

Hint: Use rightmost set bit to split and find unique numbers [OK]

Common Mistakes:

• Mixing up repeating and unique elements
• Not splitting array by set bit
• Incorrect XOR calculations