Two Non Repeating Elements in Array Using XOR in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the XOR of all elements in the array.

DSA Go

xor := 0
for _, num := range arr {
    xor = xor [1] num
}

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

A&

B^

C+

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition (+) instead of XOR (^).

Using bitwise AND (&) or OR (|) which do not help find unique elements.

XOR (^) is used to find the combined XOR of all elements, which helps identify the two unique numbers.

2fill in blank

medium

Complete the code to find the rightmost set bit in the XOR result.

DSA Go

rightmostSetBit := xor & -[1]

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

A1

B2

C0

Dxor

Attempts:

3 left

💡 Hint

Common Mistakes

Using a constant like 1 or 2 instead of xor.

Using bitwise OR instead of AND.

The rightmost set bit is found by ANDing xor with its two's complement (-xor).

3fill in blank

hard

Fix the error in the loop that separates elements into two groups based on the rightmost set bit.

DSA Go

for _, num := range arr {
    if num & [1] != 0 {
        res1 = res1 ^ num
    } else {
        res2 = res2 ^ num
    }
}

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

ArightmostSetBit

Bxor

Cnum

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using xor instead of rightmostSetBit in the condition.

Using a constant 1 which may not isolate the correct bit.

We use rightmostSetBit to check which group each number belongs to.

4fill in blank

hard

Fill both blanks to initialize the result variables and print the two unique numbers.

DSA Go

res1, res2 := [1], [2]
// After processing
fmt.Println("Two unique numbers are:", res1, res2)

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

A0

B1

C-1

Dxor

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with xor or -1 which causes wrong results.

Using 1 which changes the XOR outcome.

Initialize res1 and res2 to 0 before XORing elements into them.

5fill in blank

hard

Fill all three blanks to complete the function that returns the two unique numbers.

DSA Go

func findTwoUnique(arr []int) (int, int) {
    xor := 0
    for _, num := range arr {
        xor = xor [1] num
    }
    rightmostSetBit := xor & -[2]
    res1, res2 := [3], 0
    for _, num := range arr {
        if num & rightmostSetBit != 0 {
            res1 = res1 ^ num
        } else {
            res2 = res2 ^ num
        }
    }
    return res1, res2
}

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

A^

Bxor

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of XOR.

Using constants instead of xor variable for two's complement.

Not initializing res1 to zero.

Use XOR (^) to combine elements, xor variable for two's complement, and initialize res1 to 0.

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