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

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 against all others.

The Problem

Manually comparing each sock with every other sock is slow and tiring. It takes a lot of time and you can easily make mistakes by missing pairs or mixing socks up.

The Solution

Using XOR, a special operation, you can cleverly find the two unique socks by comparing all socks together. XOR cancels out pairs automatically, leaving only the unique socks to be found easily.

Before vs After

✗ Before

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

✓ After

xor := 0
for _, num := range 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 quickly and without extra memory.

Real Life Example

Finding two unique defective products in a batch where all others are identical pairs, without checking each product twice.

Key Takeaways

Manual pair checking is slow and error-prone.

XOR operation cancels pairs and isolates unique elements.

Efficiently finds two unique elements with simple steps.

Practice

1. What is the main idea behind using XOR to find two non-repeating elements in an array where every other element repeats?

easy

A. XOR of all elements gives XOR of the two unique elements

B. XOR sorts the array to separate unique elements

C. XOR counts the frequency of each element

D. XOR removes duplicates by summing elements

Solution

1. Step 1: Understand XOR properties

   XOR of a number with itself is 0, and XOR with 0 is the number itself.

2. Step 2: Apply XOR to all elements

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

3. Final Answer:

   XOR of all elements gives XOR of the two unique elements -> Option A

4. Quick Check:

   XOR cancels duplicates, leaving unique pair XOR [OK]

Hint: XOR all elements to isolate unique pair XOR quickly [OK]

Common Mistakes:

• Thinking XOR sorts or counts elements
• Assuming XOR gives unique elements directly
• Confusing XOR with addition or subtraction

2. Which of the following Go code snippets correctly initializes the XOR variable to start finding two non-repeating elements?

easy

A. xor := nil

B. xor := 0

C. xor := 1

D. xor := []int{}

Solution

1. Step 1: Identify correct XOR initialization

   XOR operation starts with 0 because XOR with 0 returns the number itself.

2. Step 2: Check Go syntax for variable initialization

   "xor := 0" correctly initializes an integer variable for XOR operations.

3. Final Answer:

   xor := 0 -> Option B

4. Quick Check:

   Start XOR with 0 for correct accumulation [OK]

Hint: Initialize XOR with 0 to start XOR operations correctly [OK]

Common Mistakes:

• Using 1 instead of 0 for XOR start
• Trying to initialize XOR with nil or slice
• Incorrect Go syntax for variable declaration

3. What is the output of the following Go code snippet? arr := []int{2, 3, 7, 9, 2, 3, 9, 11} xor := 0 for _, v := range arr { xor ^= v } fmt.Println(xor)

medium

A. 12

B. 18

C. 0

D. 11

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, 14^9=7, 7^11=12.

2. Step 2: Verify final XOR value

   The final XOR value after processing all elements is 12.

3. Final Answer:

   12 -> Option A

4. Quick Check:

   XOR of all elements = 12 [OK]

Hint: XOR duplicates cancel; XOR unique pair remains [OK]

Common Mistakes:

• Miscomputing XOR stepwise
• Assuming XOR sums elements
• Ignoring duplicates canceling out

4. Identify the bug in this Go code snippet that tries to find two non-repeating elements using XOR: arr := []int{4, 1, 2, 1, 2, 5} xor := 0 for i := 0; i <= len(arr); i++ { xor ^= arr[i] } fmt.Println(xor)

medium

A. Use a map instead of XOR for this problem

B. XOR variable should be initialized to 1

C. Loop condition should be i < len(arr), not i <= len(arr)

D. Array elements should be converted to strings before XOR

Solution

1. Step 1: Check loop boundary

   Loop uses i <= len(arr), which causes out-of-range error because valid indices are 0 to len(arr)-1.

2. Step 2: Correct loop condition

   Change to i < len(arr) to avoid index out of range and correctly XOR all elements.

3. Final Answer:

   Loop condition should be i < len(arr), not i <= len(arr) -> Option C

4. Quick Check:

   Index out of range fixed by correct loop boundary [OK]

Hint: Use i < len(arr) to avoid index errors in loops [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Initializing XOR with wrong value
• Misunderstanding XOR usage for strings

5. Given the array arr := []int{6, 3, 5, 4, 5, 3, 4, 7}, which two numbers are the non-repeating elements found using the XOR method?

hard

A. [5, 6]

B. [3, 5]

C. [4, 7]

D. [6, 7]

Solution

1. Step 1: XOR all elements to get xorVal

   Calculate XOR: 6^3=5, 5^5=0, 0^4=4, 4^5=1, 1^3=2, 2^4=6, 6^7=1. Final xorVal = 1.

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

   Rightmost set bit in xorVal=1. Partition array into two groups based on this bit and XOR separately to find unique elements 6 and 7.

3. Final Answer:

   [6, 7] -> Option D

4. Quick Check:

   Unique elements are 6 and 7 after XOR partition [OK]

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

Common Mistakes:

• Not partitioning array by set bit
• Mixing up duplicates with unique elements
• Incorrect XOR calculations