human
Maximize length of Subarray of 1 's after removal of a pair of consecutive Array elements | Function to find the maximum subarray length of ones.
Stores the length , starting index and ending index of the subarrays.
S : starting index of the sub - array.
Traverse only continuous 1 s.
Calculate length of the sub - array.
v [ i ] [ 0 ] : Length of subarray v [ i ] [ 1 ] : Starting Index of subarray v [ i ] [ 2 ] : Ending Index of subarray.
If no such sub - array exists.
Traversing through the subarrays.
Update maximum length.
v [ i + 1 ] [ 1 ] - v [ i ] [ 2 ] - 1 : Count of zeros between the two sub - arrays.
Update length of both subarrays to the maximum.
Update length of both subarrays - 1 to the maximum.
Check if the last subarray has the maximum length.
Driver Code"
Solution in Python:
 ... Give me a code to maximize length of Subarray of 1 's after removal of a pair of consecutive Array elements and the other steps required?
assistant
#Here is code to maximize length of Subarray of 1 's after removal of a pair of consecutive Array elements and the other steps required.
 def maxLen ( A , N ) :
	v = [ ]
	i = 0
	while i < N :
		if ( A [ i ] == 1 ) :
			s = i
			while ( i < N and A [ i ] == 1 ) :
				i += 1
			le = i - s
			v . append ( [ le , s , i - 1 ] )
		i += 1
	if ( len ( v ) == 0 ) :
		return - 1
	ans = 0
	for i in range ( len ( v ) - 1 ) :
		ans = max ( ans , v [ i ] [ 0 ] )
		if ( v [ i + 1 ] [ 1 ] - v [ i ] [ 2 ] - 1 == 2 ) :
			ans = max ( ans , v [ i ] [ 0 ] + v [ i + 1 ] [ 0 ] )
		if ( v [ i + 1 ] [ 1 ] - v [ i ] [ 2 ] - 1 == 1 ) :
			ans = max ( ans , v [ i ] [ 0 ] + v [ i + 1 ] [ 0 ] - 1 )
	ans = max ( v [ len ( v ) - 1 ] [ 0 ] , ans )
	return ans
if __name__ == " _ _ main _ _ " :
	arr = [ 1 , 0 , 1 , 0 , 0 , 1 ]
	N = len ( arr )
	print ( maxLen ( arr , N ) )