LeetCode 605. Can Place Flowers

 605. Can Place Flowers

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.

 

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 * 104
• flowerbed[i] is 0 or 1.
• There are no two adjacent flowers in flowerbed.
• 0 <= n <= flowerbed.length

Solution :-

Two Solution :-

1) By modifying vectore :

	class Solution {
	public:
		bool canPlaceFlowers(vector<int>& flowerbed, int n) {
    
    flowerbed.insert(flowerbed.begin(),0);
    flowerbed.push_back(0);
    
    for(int i = 1; i < flowerbed.size()-1; ++i)
    {
        if(flowerbed[i-1] + flowerbed[i] + flowerbed[i+1] == 0)
        {
            n--;
            ++i;
        }
            
    }
    return n <=0;
}
};

2) With out modifying vector :

class Solution {
public:
	bool canPlaceFlowers(vector<int>& flowerbed, int n) {

	   int m = flowerbed.size() , count=0 , i=0;

	while(i<m-1)
	{

	//checking two adjacent position and maintaining variable i accordingly 
    if(!flowerbed[i] && !flowerbed[i+1]){
        count++;
        i += 2;
    }
    else if(flowerbed[i]==1)
        i += 2;
    else if(flowerbed[i+1]==1)
        i += 3;
		
}
    
// checking for last index in case n is odd
if(i<m && flowerbed[i]==0)
    count++;

    
return count>=n;
    
}
};

Time Complexity :- O(N)
Space Complexity - O(1)