Find the Only Non Repeating Element Using XOR in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of XOR to find non-repeating element

What is the output of the following Go code that uses XOR to find the only non-repeating element in the array?

DSA Go

package main
import "fmt"
func main() {
  arr := []int{2, 3, 5, 4, 5, 3, 4}
  result := 0
  for _, v := range arr {
    result ^= v
  }
  fmt.Println(result)
}

A 2

B 0

C 5

D 4

💡 Hint

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

All repeating numbers cancel out because XOR of a number with itself is zero. Only the non-repeating number remains.

❓ Predict Output

intermediate

Result of XOR with multiple pairs and one unique

What will be printed by this Go program that finds the unique element using XOR?

DSA Go

package main
import "fmt"
func main() {
  nums := []int{10, 14, 10, 14, 20}
  unique := 0
  for _, num := range nums {
    unique ^= num
  }
  fmt.Println(unique)
}

A 14

B 10

C 0

D 20

💡 Hint

Pairs cancel out, leaving the unique number.

10 XOR 10 = 0, 14 XOR 14 = 0, so only 20 remains after XOR.

🧠 Conceptual

advanced

Why XOR helps find the unique element?

Why does XOR operation help in finding the only non-repeating element in an array where every other element repeats twice?

A Because XOR multiplies all numbers and divides by the repeating ones

B Because XOR adds all numbers together to find the unique sum

C Because XOR of a number with itself is zero and XOR with zero is the number itself

D Because XOR sorts the array and picks the middle element

💡 Hint

Think about what happens when you XOR a number twice.

XOR cancels out pairs because n ^ n = 0, and 0 ^ n = n, so only the unique number remains.

❓ Predict Output

advanced

Output when array has multiple unique elements

What will be the output of this Go code that uses XOR on an array with two unique elements?

DSA Go

package main
import "fmt"
func main() {
  arr := []int{1, 2, 3, 2, 1, 4}
  res := 0
  for _, val := range arr {
    res ^= val
  }
  fmt.Println(res)
}

A 7

B 4

C 3

D 0

💡 Hint

XOR of two unique numbers is their combined XOR value.

1 and 2 appear twice and cancel out. XOR of 3 and 4 is 7, so output is 7.

🔧 Debug

expert

Identify the error in XOR unique element code

What error will this Go code produce when trying to find the unique element using XOR?

DSA Go

package main
import "fmt"
func main() {
  arr := []int{5, 7, 5, 9, 7}
  var result int
  for i := 0; i < len(arr); i++ {
    result ^= arr[i]
  }
  fmt.Println(result)
}

A compilation error: missing import

B runtime error: index out of range

C prints 9

D prints 0

💡 Hint

Check the loop boundary condition carefully.

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

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