Find the Only Non Repeating Element Using XOR in DSA Go - Build from Scratch

Find the Only Non Repeating Element Using XOR

📖 Scenario: Imagine you have a list of product IDs scanned at a store. Every product ID appears twice except one product that was scanned only once by mistake. You want to find that unique product ID quickly.

🎯 Goal: Build a Go program that finds the only non-repeating element in a list of integers using the XOR operation.

📋 What You'll Learn

Create a slice of integers called products with the exact values: 4, 1, 2, 1, 2

Create a variable called unique and initialize it to 0

Use a for loop with variable product to iterate over products

Inside the loop, update unique by XORing it with product

Print the value of unique to display the only non-repeating element

💡 Why This Matters

🌍 Real World

Finding unique elements quickly is useful in error detection, data cleaning, and identifying anomalies in data streams.

💼 Career

Understanding bitwise operations like XOR helps in optimizing algorithms and solving problems efficiently in software development and data engineering roles.

1

Create the list of product IDs

Create a slice of integers called products with the exact values: 4, 1, 2, 1, 2

DSA Go

# Your code here

Hint

Use products := []int{4, 1, 2, 1, 2} to create the slice.

2

Initialize the variable to hold the unique product

Create a variable called unique and initialize it to 0

DSA Go

package main

func main() {
    products := []int{4, 1, 2, 1, 2}
    // Your code here
}

Hint

Use unique := 0 to start with zero.

3

Use XOR to find the unique product

Use a for loop with variable product to iterate over products. Inside the loop, update unique by XORing it with product using unique ^= product

DSA Go

package main

func main() {
    products := []int{4, 1, 2, 1, 2}
    unique := 0
    // Your code here
}

Hint

Use for _, product := range products and inside the loop write unique ^= product.

4

Print the unique product ID

Write fmt.Println(unique) to display the only non-repeating product ID. Remember to import fmt at the top.

DSA Go

package main

func main() {
    products := []int{4, 1, 2, 1, 2}
    unique := 0
    for _, product := range products {
        unique ^= product
    }
    // Your code here
}

Hint

Use fmt.Println(unique) to print the result. Don't forget to add import "fmt" at the top.

Practice

1. What does the XOR operation do when applied between two identical numbers?

easy

A. It returns the sum of the numbers

B. It returns 0

C. It returns 1

D. It returns the number itself

Solution

1. Step 1: Understand XOR property for identical numbers

   XOR of a number with itself always results in 0 because bits cancel out.

2. Step 2: Apply this property to the question

   Since the two identical numbers cancel out, the result is 0.

3. Final Answer:

   It returns 0 -> Option B

4. Quick Check:

   n XOR n = 0 [OK]

Hint: XOR of same numbers always zeroes out [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR operation
• Assuming XOR sums numbers

2. Which of the following Go code snippets correctly initializes a variable to store XOR results for an integer slice named nums?

easy

A. var result int = 0

B. result := []int{}

C. var result string = ""

D. result := make(map[int]int)

Solution

1. Step 1: Identify the correct type for XOR accumulation

   XOR operates on integers, so the variable must be an integer initialized to 0.

2. Step 2: Check each option

   var result int = 0 initializes an integer variable to 0, suitable for XOR operations. Others use wrong types or data structures.

3. Final Answer:

   var result int = 0 -> Option A

4. Quick Check:

   Integer zero initialization for XOR [OK]

Hint: XOR needs an integer variable initialized to zero [OK]

Common Mistakes:

• Using slice or map instead of int
• Initializing with empty string
• Not initializing variable

3. What is the output of the following Go code?

package main
import "fmt"
func main() {
 nums := []int{2, 3, 2, 4, 3}
 result := 0
 for _, num := range nums {
  result ^= num
 }
 fmt.Println(result)
}

medium

A. 2

B. 3

C. 4

D. 0

Solution

1. Step 1: XOR all elements in the slice

   Calculate XOR stepwise: 0^2=2, 2^3=1, 1^2=3, 3^4=7, 7^3=4.

2. Step 2: Identify the non-repeating element

   The final XOR result is 4, which is the only number appearing once.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   XOR cancels duplicates, leaves 4 [OK]

Hint: XOR duplicates cancel, leftover is unique [OK]

Common Mistakes:

• Adding numbers instead of XOR
• Miscomputing XOR steps
• Ignoring duplicates

4. Identify the error in this Go code snippet that tries to find the only non-repeating element using XOR:

package main
import "fmt"
func main() {
 nums := []int{5, 1, 5, 2, 1}
 var result int
 for i := 0; i <= len(nums); i++ {
  result ^= nums[i]
 }
 fmt.Println(result)
}

medium

A. Loop condition should be i < len(nums), not i <= len(nums)

B. Variable 'result' should be initialized to 1 instead of 0

C. The XOR operator '^=' is invalid in Go

D. The slice 'nums' should be declared as an array

Solution

1. Step 1: Check loop boundary condition

   The loop runs from i=0 to i<=len(nums), which causes out-of-range access at i=len(nums).

2. Step 2: Correct the loop condition

   Change condition to i < len(nums) to avoid index out of range error.

3. Final Answer:

   Loop condition should be i < len(nums), not i <= len(nums) -> Option A

4. Quick Check:

   Index out of range fixed by loop condition [OK]

Hint: Loop index must be less than length, not equal [OK]

Common Mistakes:

• Using <= instead of <
• Initializing result incorrectly
• Misunderstanding XOR operator

5. Given an array where every element appears twice except one element which appears once, how can you find the unique element using XOR in Go? Consider the array nums := []int{7, 9, 7, 8, 9, 10, 10}. What is the unique element?

hard

A. 7

B. 9

C. 10

D. 8

Solution

1. Step 1: XOR all elements in the array

   Calculate XOR stepwise: 0^7=7, 7^9=14, 14^7=9, 9^8=1, 1^9=8, 8^10=2, 2^10=8.

2. Step 2: Identify the unique element

   The final XOR result is 8, which appears only once in the array.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   Duplicates cancel, unique is 8 [OK]

Hint: XOR all elements; duplicates vanish, unique remains [OK]

Common Mistakes:

• Adding numbers instead of XOR
• Forgetting to initialize result to zero
• Misreading array elements