human
Optimally accommodate 0 s and 1 s from a Binary String into K buckets | Function to find the minimum required sum using dynamic programming.
If both start and bucket reached end then return 0 else that arrangement is not possible so return INT_MAX.
Corner case.
If state if already calculated then return its answer.
Start filling zeroes and ones which to be accommodated in jth bucket then ans for current arrangement will be ones * zeroes + recur ( i + 1 , bucket + 1 ).
If this arrangement is not possible then don 't calculate further.
Function to initialize the dp and call solveUtil ( ) method to get the answer.
Start with 0 - th index and 1 bucket.
Driver code.
K buckets"
Solution in Python:

Give me a code to optimally accommodate 0 s and 1 s from a Binary String into K buckets and the other steps required?
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#Here is code to optimally accommodate 0 s and 1 s from a Binary String into K buckets and the other steps required.
 def solveUtil ( start , bucket , str1 , K , dp ) :
	N = len ( str1 )
	if ( start == N ) :
		if ( bucket == K ) :
			return 0
		return 10 ** 9
	if ( bucket == K ) :
		return 10 ** 9
	if ( dp [ start ] [ bucket ] != - 1 ) :
		return dp [ start ] [ bucket ]
	zeroes = 0
	ones = 0
	ans = 10 ** 9
	for i in range ( start , N ) :
		if ( str1 [ i ] == '1' ) :
			ones += 1
		else :
			zeroes += 1
		if ( ones * zeroes > ans ) :
			break
		temp = solveUtil ( i + 1 , bucket + 1 , str1 , K , dp )
		if ( temp != 10 ** 9 ) :
			ans = min ( ans , temp + ( ones * zeroes ) )
	dp [ start ] [ bucket ] = ans
	return ans
def solve ( str1 , K ) :
	N = len ( str1 )
	dp = [ [ - 1 for i in range ( K ) ] for i in range ( N ) ]
	ans = solveUtil ( 0 , 0 , str1 , K , dp )
	if ans == 10 ** 9 :
		return - 1
	else :
		return ans
s = "0101"
S = [ i for i in s ]
K = 2
print ( solve ( S , K ) )