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Find four missing numbers in an array containing elements from 1 to N | Finds missing 4 numbers in O ( N ) time and O ( 1 ) auxiliary space ..
To keep track of 4 possible numbers greater than length of input array In Java , helper is automatically initialized as 0..
Traverse the input array and mark visited elements either by marking them as negative in arr [ ] or in helper [ ] ..
If element is smaller than or equal to length , mark its presence in arr [ ].
Mark presence in helper [ ].
Print all those elements whose presence is not marked ..
Driver code"
Solution in Python:
assistant
#Here is code to find four missing numbers in an array containing elements from 1 to N and the other steps required.
 def missing4 ( arr ) :
	helper = [ 0 ] * 4
	for i in range ( 0 , len ( arr ) ) :
		temp = abs ( arr [ i ] )
		if ( temp <= len ( arr ) ) :
			arr [ temp - 1 ] = arr [ temp - 1 ] * ( - 1 )
		elif ( temp > len ( arr ) ) :
			if ( temp % len ( arr ) ) :
				helper [ temp % len ( arr ) - 1 ] = - 1
			else :
				helper [ ( temp % len ( arr ) ) + len ( arr ) - 1 ] = - 1
	for i in range ( 0 , len ( arr ) ) :
		if ( arr [ i ] > 0 ) :
			print ( ( i + 1 ) , end = " ▁ " )
	for i in range ( 0 , len ( helper ) ) :
		if ( helper [ i ] >= 0 ) :
			print ( ( len ( arr ) + i + 1 ) , end = " ▁ " )
arr = [ 1 , 7 , 3 , 12 , 5 , 10 , 8 , 4 , 9 ]
missing4 ( arr )