Formatted question description: https://leetcode.ca/all/605.html

# 605. Can Place Flowers (Easy)

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.

Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1 means not empty, and an integer n, return if n new flowers can be planted in the flowerbed 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


Constraints:

• 1 <= flowerbed.length <= 2 * 104
• flowerbed[i] is 0 or 1.
• There are no two adjacent flowers in flowerbed.
• 0 <= n <= flowerbed.length

Related Topics:
Array, Greedy

Similar Questions:

## Solution 1.

A placeable plot should be empty, as well as its direct neighbors, if any. So, go through the flowerbed, whenever you find a placeable plot, place the flower into it, decrement n. Finally, return n == 0.

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

// Time: O(N)
// Space: O(N)
class Solution {
public:
bool canPlaceFlowers(vector<int>& A, int n) {
for (int i = 0; i < A.size() && n > 0; ++i) {
if (A[i] == 1) continue;
if ((i == 0 || A[i - 1] == 0) && (i == A.size() - 1 || A[i + 1] == 0)) {
A[i] = 1;
--n;
}
}
return n == 0;
}
};


Java

• class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
List<Integer> emptyCounts = new ArrayList<Integer>();
int prevEmpty = -1;
int emptyCount = 1;
int length = flowerbed.length;
for (int i = 0; i <= length; i++) {
int num = i < length ? flowerbed[i] : 0;
if (num == 0) {
emptyCount++;
prevEmpty = i;
} else {
if (emptyCount > 0) {
emptyCount = 0;
}
}
}
if (emptyCount > 0)
int maxFlowers = 0;
for (int num : emptyCounts)
maxFlowers += (num - 1) / 2;
return maxFlowers >= n;
}
}

• // OJ: https://leetcode.com/problems/can-place-flowers
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool canPlaceFlowers(vector<int>& A, int n) {
for (int i = 0; i < A.size() && n > 0; ++i) {
if (A[i] == 1) continue;
if ((i == 0 || A[i - 1] == 0) && (i == A.size() - 1 || A[i + 1] == 0)) {
A[i] = 1;
--n;
}
}
return n == 0;
}
};

• class Solution(object):
def canPlaceFlowers(self, flowerbed, n):
"""
:type flowerbed: List[int]
:type n: int
:rtype: bool
"""
ans = 0
cnt = 1
for plot in flowerbed:
if plot == 0:
cnt += 1
else:
ans += abs(cnt - 1) / 2
cnt = 0
return ans + cnt / 2 >= n