Can place flowers

Time: O(N); Space: O(1); easy

Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots - they would compete for water and both would die.

Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a number N, return if N new flowers can be planted in it without violating the no-adjacent-flowers rule.

Example 1:

Input: flowerbed = [1,0,0,0,1], n = 1

Output: True

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2

Output: False

Notes:

• The input array won’t violate no-adjacent-flowers rule.

• The input array size is in the range of [1, 20000].

• N is a non-negative integer which won’t exceed the input array size.

class Solution1(object):
    def canPlaceFlowers(self, flowerbed, n):
        """
        :type flowerbed: List[int]
        :type n: int
        :rtype: bool
        """
        for i in range(len(flowerbed)):
            if flowerbed[i] == 0 and (i == 0 or flowerbed[i-1] == 0) and \
                (i == len(flowerbed)-1 or flowerbed[i+1] == 0):
                flowerbed[i] = 1
                n -= 1
            if n <= 0:
                return True
        return False

s = Solution1()
flowerbed = [1,0,0,0,1]
N = 1
assert s.canPlaceFlowers(flowerbed, N) == True
N = 2
assert s.canPlaceFlowers(flowerbed, N) == False

See also:

https://leetcode.com/problems/can-place-flowers