missingAndRepeatingNumberPython

def getTwoElements(arr, n):      
    x = 0
    y = 0

    # xor1 will hold xor of all elements and numbers from 1 to n
    xor1 = arr[0]

    # Get the xor of all array elements 
    for i in range(1, n):
        xor1 = xor1 ^ arr[i]

    # XOR the previous result with numbers from 1 to n
    for i in range(1, n + 1):
        xor1 = xor1 ^ i
    
    # set_bit_no will have only single set bit of xor1 
    # Get the rightmost set bit in set_bit_no 
    set_bit_no = xor1 & ~(xor1 - 1)

    # Now divide elements into two sets by comparing rightmost set bit of xor1 with the bit at the same 
    # position in each element. Also, get XORs of two sets.
    # The two XORs are the output elements. The following two for loops serve the purpose
    
    for i in range(n):
        if ((arr[i] & set_bit_no) != 0):
            # arr[i] belongs to first set
            x = x ^ arr[i]

        else:
            # arr[i] belongs to second set
            y = y ^ arr[i]

    for (i = 1 i <= n i++):
        if ((i & set_bit_no) != 0):
            # i belongs to first set
            x = x ^ i

        else:
            # i belongs to second set
            y = y ^ i

    
    # at last do a linear check which amont x and y is missing or repeating 

    # *x and *y hold the desired output elements */