Description#
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 * 104flowerbed[i] is 0 or 1.- There are no two adjacent flowers in
flowerbed. 0 <= n <= flowerbed.length
Solutions#
Solution 1: Greedy#
We directly traverse the array $flowerbed$. For each position $i$, if $flowerbed[i]=0$ and its adjacent positions on the left and right are also $0$, then we can plant a flower at this position. Otherwise, we cannot. Finally, we count the number of flowers that can be planted. If it is not less than $n$, we return $true$, otherwise we return $false$.
The time complexity is $O(n)$, where $n$ is the length of the array $flowerbed$. We only need to traverse the array once. The space complexity is $O(1)$.
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| class Solution:
def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
flowerbed = [0] + flowerbed + [0]
for i in range(1, len(flowerbed) - 1):
if sum(flowerbed[i - 1 : i + 2]) == 0:
flowerbed[i] = 1
n -= 1
return n <= 0
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| class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int m = flowerbed.length;
for (int i = 0; i < m; ++i) {
int l = i == 0 ? 0 : flowerbed[i - 1];
int r = i == m - 1 ? 0 : flowerbed[i + 1];
if (l + flowerbed[i] + r == 0) {
flowerbed[i] = 1;
--n;
}
}
return n <= 0;
}
}
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| class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
int m = flowerbed.size();
for (int i = 0; i < m; ++i) {
int l = i == 0 ? 0 : flowerbed[i - 1];
int r = i == m - 1 ? 0 : flowerbed[i + 1];
if (l + flowerbed[i] + r == 0) {
flowerbed[i] = 1;
--n;
}
}
return n <= 0;
}
};
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| func canPlaceFlowers(flowerbed []int, n int) bool {
m := len(flowerbed)
for i, v := range flowerbed {
l, r := 0, 0
if i > 0 {
l = flowerbed[i-1]
}
if i < m-1 {
r = flowerbed[i+1]
}
if l+v+r == 0 {
flowerbed[i] = 1
n--
}
}
return n <= 0
}
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| function canPlaceFlowers(flowerbed: number[], n: number): boolean {
const m = flowerbed.length;
for (let i = 0; i < m; ++i) {
const l = i === 0 ? 0 : flowerbed[i - 1];
const r = i === m - 1 ? 0 : flowerbed[i + 1];
if (l + flowerbed[i] + r === 0) {
flowerbed[i] = 1;
--n;
}
}
return n <= 0;
}
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| impl Solution {
pub fn can_place_flowers(flowerbed: Vec<i32>, n: i32) -> bool {
let (mut flowers, mut cnt) = (vec![0], 0);
flowers.append(&mut flowerbed.clone());
flowers.push(0);
for i in 1..flowers.len() - 1 {
let (l, r) = (flowers[i - 1], flowers[i + 1]);
if l + flowers[i] + r == 0 {
flowers[i] = 1;
cnt += 1;
}
}
cnt >= n
}
}
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| class Solution {
/**
* @param Integer[] $flowerbed
* @param Integer $n
* @return Boolean
*/
function canPlaceFlowers($flowerbed, $n) {
array_push($flowerbed, 0);
array_unshift($flowerbed, 0);
for ($i = 1; $i < count($flowerbed) - 1; $i++) {
if ($flowerbed[$i] === 0) {
if ($flowerbed[$i - 1] === 0 && $flowerbed[$i + 1] === 0) {
$flowerbed[$i] = 1;
$n--;
}
}
}
return $n <= 0;
}
}
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