CareerCup

It's a dynamic programming question. And the solution format is just like the fabocci sequence. For example, if w = 1, then the answer is 1, if w = 2, the answer is 2, if w = 3, the answer is 1 + 2 = 3, if w = 4, the answer is 2 + 3 = 5. And the dynamic equation is
dp[n] = dp[n - 1] + dp[n - 2]. Here is the code:

public class Solution{
    public int dp(int w){
        if(w == 1){
            return 1;
        }
        if(w == 2){
            return 2;
        }
        int prev1 = 1;
        int prev2 = 2;
        int current = 0;
        for(int i = 3; i <= w; ++i){
            current = prev1 + prev2;
            prev1 = prev2;
            prev2 = current;
        }
        return current;
    }

    public static void main(String[] args){
        int w = 4;
        Solution solution = new Solution();
        System.out.println(solution.dp(w));
    }
}

Above problem is nothing but a fibbonacci series .
f(n) = f(n-1) + f(n-2)

@striker, the question isn't how many tiles, the question is how many ways.

no, it seems to me that it is not fibonacci problem, it is a simple probability problem

for eg, for 3 = 3, using ur fiboinacci formula, it is giving answer as 3, but there are only 2 ways we can put tiles on a 2 * 3 floor.

the answer is (W/2) * 2 ways in total.

Please correct if my understanding of the question is wrong.

Isn't the answer to this N!
when w=1 there is 1 way
when w=2 there is 2 way
when w =3 there is 6 ways
just think of the tiles as a,b,c in the 3 slots it can be arranged in 6 different ways

The number of tiles will be equal to the value "W".

Need more details on this. For any value of "W", there will be only one (1) way to do this, as we have all tiles of same dimension (2 x 1).