Can Place Flowers

Description

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:

1. The input array won't violate no-adjacent-flowers rule.

2. The input array size is in the range of [1, 20000].

3. n is a non-negative integer which won't exceed the input array size.

Solutions

My solution

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int size = flowerbed.size();
        int i = 0;
        while(i < size && flowerbed[i] == 0) ++i;
        n -= i == size ? (i + 1) / 2 : i / 2;
        while(i < size){
            int j = i + 1;
            while(j < size && flowerbed[j] == 0) ++j;
            n -= j == size ? (j - i - 1) / 2 : (j - i - 2) / 2;
            if(n <= 0) return true;
            i = j;
        }
        return n <= 0;
    }
};

Optimal solution

Credit: https://leetcode.com/problems/can-place-flowers/discuss/103883/Java-Very-easy-solution

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int zeros = 1; // the essential part: setting # of zeros to 1
        for(int num : flowerbed){
            if(num == 0){
                ++zeros;
            }else{
                n -= (zeros - 1) / 2;
                if(n <= 0) return true;
                zeros = 0;
            }
        }
        n -= zeros / 2;
        return n <= 0;
    }
};