Solution to beads problem

Beads.py

"""
Beads problem. Written by Andati Rodgers on 6th Nov, 2012
Used aspects of dynamic programming

"""
def length(s): #determine possible max length of b's followed by r's given a string
	s2 =s
	b=r=pr=mx=index=i=j=0
	same = True
	for k in range(0,len(s)-1):
		if s2[k] != s2[k+1]:
			same = False
			break
	if same == True:
		return len(s2),0
	while j<len(s2)*2:
		if i==len(s2):
			i=0
		if s2[i] == 'b':
			if s2[i-1]=='r':
				if mx < b+r:
					mx = b+r; index = i-r
				b=0; pr=r; r=0				
			b+=1
		elif s2[i] == 'r':
			if s2[i-1]=='b':
				if mx < b+pr:
					mx = b+pr; index = i-b
			r+=1
		i+=1; j+=1
	if index<0:
		index = len(s2)+index
	return mx,index

def dynamic(s,i): # dynamically look for all possible combinations of letter 'w' - if assigned r or b
	if i>=len(s):
		global mx,index
		num2,index2 = length(s)
		if num2 > mx:
			mx = num2; index = index2
		return
	ss = list(s)
	if ss[i]=='w':
		ss[i]='r'
		dynamic("".join(ss),i+1)
		ss[i]='b'
		dynamic("".join(ss),i+1)
	else:
		dynamic("".join(ss),i+1)

if __name__=="__main__":
	#ss = "wwwbbrwrbrbrrbrbrwrwwrbwrwrrb"
	ss = "bbb"
	mx = index= 0
	dynamic(ss,0)
	print "Length",mx, "Index", index