Why Two Non Repeating Elements in Array Using XOR in DSA Javascript?

The Big Idea

Discover how a simple XOR trick can instantly spot two unique items hidden in a crowd!

The Scenario

Imagine you have a big box of socks where every sock has a matching pair except for two unique socks. You want to find those two unique socks quickly without checking every single sock against all others.

The Problem

Manually comparing each sock with every other sock takes a lot of time and effort. It's easy to miss pairs or get confused, especially if the box is very large. This slow and error-prone method wastes your time and patience.

The Solution

Using XOR, a special operation, you can cleverly find the two unique socks by combining all socks in a way that pairs cancel out. This leaves you with a simple way to separate and identify the two unique socks quickly and without mistakes.

Before vs After

✗ Before

for(let i=0; i<arr.length; i++) {
  for(let j=i+1; j<arr.length; j++) {
    if(arr[i] === arr[j]) {
      // mark as paired
    }
  }
}
// find unmarked elements

✓ After

let xor = 0;
for(let num of arr) xor ^= num;
// use xor to find two unique numbers

What It Enables

This method lets you find two unique elements in a large list instantly, saving time and avoiding confusion.

Real Life Example

In a warehouse, quickly spotting two unique defective items among thousands of identical products without checking each one manually.

Key Takeaways

Manual pair checking is slow and error-prone.

XOR operation cancels pairs and isolates unique elements.

Efficiently finds two unique elements in one pass.

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