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.
- The code is self explanatory.
- Time Complexity: O(N)
- Space Complexity: O(1)
class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n)
{
int m = flowerbed.size();
if (m == 1 && flowerbed[0] == 0) // if only one flowerbed and it is empty
return n <= 1;
for (int i = 0; i < m; i++) {
if (i == 0) { // first element
if (flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
flowerbed[i] = 1;
--n;
}
} else if (i == m - 1) { // last element
if (flowerbed[i - 1] == 0 && flowerbed[i] == 0) {
flowerbed[i] = 1;
--n;
}
} else { // other elements
if (flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
flowerbed[i] = 1;
--n;
}
}
}
return n <= 0;
}
};
- Solution by using
pre
for previous element andnext
for next element. - Time Complexity: O(N)
- Space Complexity: O(1)
class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n)
{
int m = flowerbed.size();
for (int i = 0; i < m; i++) {
if (flowerbed[i] == 0) {
int prev = (i == 0) ? 0 : flowerbed[i - 1];
int next = (i == m - 1) ? 0 : flowerbed[i + 1];
if (next == 0 && prev == 0) {
flowerbed[i] = 1;
--n;
}
}
}
return n <= 0;
}
};