human
Fill array with 1 's using minimum iterations of filling neighbors | Returns count of iterations to fill arr [ ] with 1 s ..
Start traversing the array.
Traverse until a 0 is found.
Count contiguous 0 s.
Condition for Case 3.
Condition to check if Case 1 satisfies :.
If count_zero is even.
If count_zero is odd.
Reset count_zero.
Case 2.
Update res.
Driver code"
How can the above be solved in Python?
 ... What is a code to fill array with 1 's using minimum iterations of filling neighbors and the other steps required?
assistant
#Here is code to fill array with 1 's using minimum iterations of filling neighbors and the other steps required.
 def countIterations ( arr , n ) :
	oneFound = False ;
	res = 0 ;
	i = 0 ;
	while ( i < n ) :
		if ( arr [ i ] == 1 ) :
			oneFound = True ;
		while ( i < n and arr [ i ] == 1 ) :
			i += 1 ;
		count_zero = 0 ;
		while ( i < n and arr [ i ] == 0 ) :
			count_zero += 1 ;
			i += 1 ;
		if ( oneFound == False and i == n ) :
			return - 1 ;
		curr_count = 0 ;
		if ( i < n and oneFound == True ) :
			if ( ( count_zero & 1 ) == 0 ) :
				curr_count = count_zero // 2 ;
			else :
				curr_count = ( count_zero + 1 ) // 2 ;
			count_zero = 0 ;
		else :
			curr_count = count_zero ;
			count_zero = 0 ;
		res = max ( res , curr_count ) ;
	return res ;
arr = [ 0 , 1 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 0 ] ;
n = len ( arr ) ;
print ( countIterations ( arr , n ) ) ;