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

The Big Idea

Discover how a simple XOR trick can find two unique items instantly in a sea of pairs!

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 one by one.

The Problem

Checking each sock manually to find the two unique ones means comparing each sock with all others. This takes a lot of time and can easily cause mistakes, especially if the box is very big.

The Solution

Using XOR, a special math trick, you can quickly find the two unique socks by combining all socks in a way that pairs cancel out, leaving only the unique ones. This saves time and effort.

Before vs After

✗ Before

for (let i = 0; i < arr.length; i++) {
  let count = 0;
  for (let j = 0; j < arr.length; j++) {
    if (arr[i] === arr[j]) count++;
  }
  if (count === 1) console.log(arr[i]);
}

✓ After

let xorAll = 0;
for (const number of arr) {
  xorAll ^= number;
}
// Further steps to find two unique numbers

What It Enables

This method enables finding two unique elements in a large collection instantly without extra memory or slow comparisons.

Real Life Example

Finding two unique defective products in a large batch where all others are identical pairs, quickly and efficiently.

Key Takeaways

Manual checking is slow and error-prone.

XOR helps cancel out pairs and isolate unique elements.

This method is fast and uses little extra space.

Practice

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

easy

A. To sort the array in ascending order

B. To count the frequency of each element

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

D. To reverse the array elements

Solution

1. Step 1: Understand XOR properties

   XOR of two same numbers is 0, and XOR with 0 is the number itself.

2. Step 2: Apply XOR on array elements

   XORing all elements cancels out repeating ones, leaving XOR of two unique elements.

3. Final Answer:

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

4. Quick Check:

   XOR cancels duplicates = isolate unique elements [OK]

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

Common Mistakes:

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

2. Which of the following TypeScript code snippets correctly initializes the XOR variable to start finding two unique elements?

easy

A. let xor = 1;

B. let xor = 0;

C. let xor = -1;

D. let xor = null;

Solution

1. Step 1: Identify initial XOR value

   XOR with 0 returns the number itself, so start with 0.

2. Step 2: Confirm initialization

   Initializing xor as 0 allows proper XOR accumulation over array elements.

3. Final Answer:

   let xor = 0; -> Option B

4. Quick Check:

   Start XOR with 0 for correct accumulation [OK]

Hint: Initialize XOR with 0 to accumulate correctly [OK]

Common Mistakes:

• Starting XOR with 1 changes results
• Using null or undefined causes errors
• Starting XOR with -1 gives wrong output

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

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

medium

A. 0

B. 9

C. 7

D. 14

Solution

1. Step 1: XOR all elements in 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 after all elements is 14.

3. Final Answer:

   14 -> Option D

4. Quick Check:

   XOR all elements = 14 [OK]

Hint: XOR all elements stepwise to find final value [OK]

Common Mistakes:

• Stopping XOR early
• Incorrect XOR calculation order
• Confusing XOR with addition

4. Identify the error in this TypeScript code snippet that tries to find two unique 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. Loop condition should be i < arr.length, not i <= arr.length

B. XOR operator should be & instead of ^

C. Variable xor should be initialized to 1

D. Array elements should be converted to strings before XOR

Solution

1. Step 1: Check loop boundary

   Loop uses i <= arr.length which causes out-of-bounds access at last iteration.

2. Step 2: Correct loop condition

   Change to i < arr.length to avoid accessing undefined element.

3. Final Answer:

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

4. Quick Check:

   Loop index must be less than array length [OK]

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

Common Mistakes:

• Using wrong loop boundary causing runtime error
• Confusing XOR with AND operator
• Initializing xor incorrectly

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

hard

A. [3, 5]

B. [6, 1]

C. [1, 3]

D. [5, 6]

Solution

1. Step 1: XOR all elements

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

2. Step 2: Find rightmost set bit of XOR

   Rightmost set bit of 6 (binary 110) is 2 (binary 010).

3. Step 3: Divide elements into two groups and XOR separately

   Group 1 (bit set): 6, 3, 6 -> XOR = 3; Group 2 (bit not set): 1, 1, 5 -> XOR = 5.

4. Final Answer:

   [3, 5] -> Option A

5. Quick Check:

   Two unique elements = [3, 5] [OK]

Hint: Use rightmost set bit to separate and XOR groups [OK]

Common Mistakes:

• Not separating elements by set bit
• Mixing order of unique elements
• Incorrect XOR calculations