human
Find the repeating and the missing | Added 3 new methods | The output of this function is stored at x and y.
Will hold xor of all elements and numbers from 1 to n.
Get the xor of all array elements.
XOR the previous result with numbers from 1 to n.
Will have only single set bit of xor1.
Now divide elements into two sets by comparing a 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.
arr [ i ] belongs to first set.
arr [ i ] belongs to second set.
i belongs to first set.
i belongs to second set.
Driver code"
How can the above be solved in Python?

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Give me a code to find the repeating and the missing and the other steps required?
assistant
#Here is code to find the repeating and the missing and the other steps required.
 def getTwoElements ( arr , n ) :
	global x , y
	x = 0
	y = 0
	xor1 = arr [ 0 ]
	for i in range ( 1 , n ) :
		xor1 = xor1 ^ arr [ i ]
	for i in range ( 1 , n + 1 ) :
		xor1 = xor1 ^ i
	set_bit_no = xor1 & ~ ( xor1 - 1 )
	for i in range ( n ) :
		if ( arr [ i ] & set_bit_no ) != 0 :
			x = x ^ arr [ i ]
		else :
			y = y ^ arr [ i ]
	for i in range ( 1 , n + 1 ) :
		if ( i & set_bit_no ) != 0 :
			x = x ^ i
		else :
			y = y ^ i
arr = [ 1 , 3 , 4 , 5 , 5 , 6 , 2 ]
n = len ( arr )
getTwoElements ( arr , n )
print ( " The ▁ missing ▁ element ▁ is " , x , " and ▁ the ▁ repeating ▁ number ▁ is " , y )