Two Non Repeating Elements in Array Using XOR in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the XOR of all elements in the array.

DSA Javascript

let xor = 0;
for (let i = 0; i < arr.length; i++) {
  xor = xor [1] arr[i];
}

Drag options to blanks, or click blank then click option'

A^

B&

C+

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition (+) instead of XOR (^).

Using bitwise AND (&) or OR (|) which do not give the desired result.

The XOR operator (^) is used to find the XOR of two numbers. Here, we XOR all elements to get the XOR of the two unique numbers.

2fill in blank

medium

Complete the code to find the rightmost set bit in xor.

DSA Javascript

let setBit = xor & -[1];

Drag options to blanks, or click blank then click option'

Aarr.length

Bxor

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using a wrong variable instead of xor.

Using 0 or 1 which do not isolate the set bit.

The expression xor & -xor isolates the rightmost set bit of xor.

3fill in blank

hard

Fix the error in the loop that separates elements into two groups based on setBit.

DSA Javascript

for (let i = 0; i < arr.length; i++) {
  if (arr[i] & [1]) {
    num1 = num1 ^ arr[i];
  } else {
    num2 = num2 ^ arr[i];
  }
}

Drag options to blanks, or click blank then click option'

Ai

Bxor

CsetBit

Dnum1

Attempts:

3 left

💡 Hint

Common Mistakes

Using xor instead of setBit in the condition.

Using the loop index i instead of the bit mask.

We use setBit to check if the bit is set in each element to separate them into two groups.

4fill in blank

hard

Fill both blanks to initialize variables and return the result array.

DSA Javascript

let num1 = [1];
let num2 = [2];
return [num1, num2];

Drag options to blanks, or click blank then click option'

A0

B1

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 1 or null which causes wrong results.

Not initializing variables before use.

Initialize num1 and num2 to 0 before XOR operations to accumulate unique numbers.

5fill in blank

hard

Fill all three blanks to complete the function that finds two non-repeating elements.

DSA Javascript

function findTwoUnique(arr) {
  let xor = 0;
  for (let i = 0; i < arr.length; i++) {
    xor = xor [1] arr[i];
  }
  let setBit = xor & -[2];
  let num1 = [3];
  let num2 = 0;
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] & setBit) {
      num1 = num1 ^ arr[i];
    } else {
      num2 = num2 ^ arr[i];
    }
  }
  return [num1, num2];
}

Drag options to blanks, or click blank then click option'

A^

Bxor

C0

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators like & or + instead of ^.

Initializing num1 incorrectly.

Using wrong variable in setBit calculation.

The function uses XOR to find combined XOR of two unique numbers, isolates a set bit, then separates numbers into two groups to find the unique elements.

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