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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

def find_xor(arr):
    result = 0
    for num in arr:
        result = result [1] num
    return result

arr = [2, 3, 7, 9, 2, 3, 9, 11]
print(find_xor(arr))

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A^

B+

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of XOR.

Using bitwise AND or OR instead of XOR.

The XOR operator ^ is used to find the XOR of two numbers. Here, we XOR all elements to get the XOR of the two unique numbers.

2fill in blank

medium

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

DSA Python

def rightmost_set_bit(xor_result):
    return xor_result & -[1]

xor_result = 12
print(rightmost_set_bit(xor_result))

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Axor_result

B1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using a constant like 1 instead of the variable.

Using bitwise OR instead of AND.

The expression xor_result & -xor_result isolates the rightmost set bit of xor_result.

3fill in blank

hard

Fix the error in the code to correctly separate the two unique numbers using the set bit.

DSA Python

def find_two_unique(arr):
    xor_all = 0
    for num in arr:
        xor_all ^= num
    set_bit = xor_all & -xor_all
    num1 = 0
    num2 = 0
    for num in arr:
        if num & [1]:
            num1 ^= num
        else:
            num2 ^= num
    return num1, num2

arr = [4, 1, 2, 1, 2, 5]
print(find_two_unique(arr))

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Anum1

Bxor_all

Cset_bit

Dnum2

Attempts:

3 left

💡 Hint

Common Mistakes

Using xor_all instead of set_bit in the if condition.

Using num1 or num2 instead of set_bit.

The variable set_bit holds the rightmost set bit used to separate the two unique numbers.

4fill in blank

hard

Fill both blanks to complete the function that returns the two unique numbers in sorted order.

DSA Python

def find_two_unique_sorted(arr):
    xor_all = 0
    for num in arr:
        xor_all ^= num
    set_bit = xor_all & -xor_all
    num1 = 0
    num2 = 0
    for num in arr:
        if num & [1]:
            num1 ^= num
        else:
            num2 ^= num
    return tuple(sorted(([2], num2)))

arr = [10, 20, 10, 30]
print(find_two_unique_sorted(arr))

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Aset_bit

Bxor_all

Cnum1

Dnum2

Attempts:

3 left

💡 Hint

Common Mistakes

Returning xor_all instead of num1.

Using xor_all in the if condition instead of set_bit.

The first blank should be set_bit to separate groups. The second blank should be num1 to return the first unique number.

5fill in blank

hard

Fill all three blanks to complete the function that finds two unique numbers using XOR and returns them sorted.

DSA Python

def two_non_repeating(arr):
    xor_all = 0
    for num in arr:
        xor_all ^= num
    set_bit = xor_all & -[1]
    num1 = 0
    num2 = 0
    for num in arr:
        if num & [2]:
            num1 ^= num
        else:
            num2 ^= num
    return tuple(sorted((num1, [3])))

arr = [1, 2, 3, 2, 1, 4]
print(two_non_repeating(arr))

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Axor_all

Bset_bit

Cnum2

Dnum1

Attempts:

3 left

💡 Hint

Common Mistakes

Using num1 instead of num2 in the return statement.

Using set_bit instead of xor_all in the negative operation.

The first blank is xor_all to get the negative for isolating the rightmost set bit. The second blank is set_bit to separate groups. The third blank is num2 to return the second unique number.