Find the Only Non Repeating Element Using XOR in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result variable to zero.

DSA Go

result := [1]

Drag options to blanks, or click blank then click option'

A-1

B1

C0

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing result to 1 or -1 causes wrong output.

Using nil causes compilation error.

We start with zero because XOR with zero leaves the number unchanged.

2fill in blank

medium

Complete the code to XOR each element with the result.

DSA Go

for _, num := range nums {
    result = result [1] num
}

Drag options to blanks, or click blank then click option'

A^

B|

C&

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or & instead of ^ changes the logic.

Using | does not cancel duplicates.

XOR operator (^) is used to find the unique element by cancelling duplicates.

3fill in blank

hard

Fix the error in the function signature to accept a slice of integers.

DSA Go

func findUnique([1] []int) int {
    result := 0
    for _, num := range [1] {
        result = result ^ num
    }
    return result
}

Drag options to blanks, or click blank then click option'

Anum

Bnums

Cnumbers

Darr

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names causes compilation errors.

Using singular 'num' as parameter name is confusing.

The parameter name must be consistent; 'nums' is used in the function body.

4fill in blank

hard

Fill both blanks to complete the main function that calls findUnique and prints the result.

DSA Go

func main() {
    nums := []int{2, 3, 5, 4, 5, 3, 4}
    unique := [1](nums)
    fmt.[2](unique)
}

Drag options to blanks, or click blank then click option'

AfindUnique

BPrintln

CPrint

DfindUniqueElement

Attempts:

3 left

💡 Hint

Common Mistakes

Using Print does not add newline.

Calling undefined function causes error.

Call findUnique to get the unique number and use fmt.Println to print it with a newline.

5fill in blank

hard

Fill all three blanks to import the required package, define the function, and print the unique element.

DSA Go

package main

import [1]

func [2](nums []int) int {
    result := 0
    for _, num := range nums {
        result = result ^ num
    }
    return result
}

func main() {
    nums := []int{1, 2, 1, 3, 2}
    unique := [3](nums)
    fmt.Println(unique)
}

Drag options to blanks, or click blank then click option'

A"fmt"

BfindUnique

D"math"

Attempts:

3 left

💡 Hint

Common Mistakes

Importing wrong package causes error.

Mismatched function names cause compilation errors.

Import fmt package, define findUnique function, and call it in main to print the unique element.

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