# Can Replace Flowers

## Problem

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
``````

Note:

• 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.

## Solution

Greedy Solution: O(n) time, O(1) space

``````public class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
if (flowerbed == null || n > flowerbed.length / 2 + 1) return false;
int count = 0;
for (int i = 0; i < flowerbed.length && count < n; i++) {
if (flowerbed[i] == 0) {
int prev = i == 0 ? 0 : flowerbed[i - 1];
int next = i == flowerbed.length - 1 ? 0 : flowerbed[i + 1];
if (prev == 0 && next == 0) {
flowerbed[i] = 1;
count++;
}
}
}
return count == n;
}
}
``````

## Analysis

This solution literally checks each position in flowerbed and increase `count` if there is available place
Therefore, we just need to return `count == 0` at the end
To check if position `i` is available, we should see its previous `i-1` and next `i+1` in given `flowerbed`
If available, we first set `flowerbed[i] = 1` then increment our `count`
To terminate the loop earlier, we add `count < n` inside for loop heading