LeetCode link: 605. Can Place Flowers, difficulty: 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 true
if n
new flowers can be planted in the flowerbed
without violating the no-adjacent-flowers rule and false
otherwise.
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 * 10^4
flowerbed[i]
is0
or1
.- There are no two adjacent flowers in
flowerbed
. 0 <= n <= flowerbed.length
Intuition
Check each empty plot (0
). If both adjacent plots are empty (or boundaries), plant a flower (set to 1
) and count. Return true
if the final count ≥ n
, otherwise false
.
Step by Step Solutions
- Initialize counter
count = 0
. - Iterate through each plot:
- If current is
1
, skip. - If current is
0
and both adjacent are0
(or boundaries), plant (1
), incrementcount
.
- If current is
- Return
count >= n
.
Complexity
Time complexity
O(N)
Space complexity
O(1)
Python #
class Solution:
def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
count = 0
for i in range(len(flowerbed)):
if flowerbed[i] == 1:
continue
if (i == 0 or flowerbed[i - 1] == 0) and \
(i == len(flowerbed) - 1 or flowerbed[i + 1] == 0):
flowerbed[i] = 1
count += 1
return count >= n
Java #
class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int count = 0;
for (int i = 0; i < flowerbed.length; i++) {
if (flowerbed[i] == 1) {
continue;
}
if ((i == 0 || flowerbed[i - 1] == 0) &&
(i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {
flowerbed[i] = 1;
count++;
}
}
return count >= n;
}
}
C++ #
class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
int count = 0;
for (int i = 0; i < flowerbed.size(); i++) {
if (flowerbed[i] == 1) {
continue;
}
if ((i == 0 || flowerbed[i - 1] == 0) &&
(i == flowerbed.size() - 1 || flowerbed[i + 1] == 0)) {
flowerbed[i] = 1;
count++;
}
}
return count >= n;
}
};
JavaScript #
var canPlaceFlowers = function(flowerbed, n) {
let count = 0;
for (let i = 0; i < flowerbed.length; i++) {
if (flowerbed[i] === 1) {
continue;
}
if ((i === 0 || flowerbed[i - 1] === 0) &&
(i === flowerbed.length - 1 || flowerbed[i + 1] === 0)) {
flowerbed[i] = 1;
count++;
}
}
return count >= n;
};
Go #
func canPlaceFlowers(flowerbed []int, n int) bool {
count := 0
for i := 0; i < len(flowerbed); i++ {
if flowerbed[i] == 1 {
continue
}
if (i == 0 || flowerbed[i - 1] == 0) &&
(i == len(flowerbed) - 1 || flowerbed[i + 1] == 0) {
flowerbed[i] = 1
count++
}
}
return count >= n
}
C# #
public class Solution
{
public bool CanPlaceFlowers(int[] flowerbed, int n)
{
int count = 0;
for (int i = 0; i < flowerbed.Length; i++)
{
if (flowerbed[i] == 1)
{
continue;
}
if ((i == 0 || flowerbed[i - 1] == 0) &&
(i == flowerbed.Length - 1 || flowerbed[i + 1] == 0))
{
flowerbed[i] = 1;
count++;
}
}
return count >= n;
}
}
Ruby #
def can_place_flowers(flowerbed, n)
count = 0
flowerbed.each_with_index do |plot, i|
next if plot == 1
if (i == 0 || flowerbed[i - 1] == 0) &&
(i == flowerbed.length - 1 || flowerbed[i + 1] == 0)
flowerbed[i] = 1
count += 1
end
end
count >= n
end