Can Place Flowers

'''
LOGIC:
In between any two ones we need to fit max 01 pairs with 0 at last.
(So 0101...01010 for odd gaps and 0101...010100 for even gaps ... Here gaps are bounded by 1)
For eg
1 _ _ _ _ _ 1.
In these five gaps the last has to be zero and then we can fit 2 01 pairs
1 _ _ _ _ 1
Here also the last gap has to be zero and in the first three we can fit only one 01 pair(the third remains zero)
So we conclude
No of pairs = (gaps-1)//2 (minus one becoz last gap has to be zero)
No. of new ones = no. of pairs
To handle leading gaps prepend 10
To handle trailing gaps append 01
'''
class Solution(object):
    def canPlaceFlowers(self, flowerbed, n):
        #flowerbed.append(0)
        #flowerbed.append(1)
        gaps = 1 #equivalent to prepending 10
        res = 0
        for i in flowerbed:
            if i==1:
                res += (gaps-1)//2
                gaps = 0
            else:
                gaps += 1
        #equivalent to appending 01
        return res+(gaps)//2 >= n