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# Program to count number of configurations are there to fill area with dominos and trominos in C++

Suppose we have two shapes, Domino and Tromino. Dominos are 2 x 1 shape and Trominos are ‘L’ like shape. They can be rotated like below −

If we have a number n, we have to find number of configurations to fill a 2 x n board with these two types of pieces. As we know in tiling, every square must be covered by a tile.

So if the input is 3, then the output will be 5. So the arrangements can be [XYZ XXZ XYY XXY XYY] and [XYZ YYZ XZZ XYY XXY], here different letters are used for different tiles.

To solve this, we will follow these steps −

Make an array called dp of size N + 5, set dp[1] := 1, dp[2] := 2 and dp[3] := 5

for i in range 4 to N

dp[i] := 2*dp[i − 1] + dp[i − 3]

return dp[N]

Let us see the following implementation to get better understanding −

## Example

#include <bits/stdc++.h> using namespace std; const int MOD = 1e9 + 7; int add(int a, int b){ return ((a % MOD) + (b % MOD)) % MOD; } class Solution { public: int solve(int N) { vector <int> dp(N + 5); dp[1] = 1; dp[2] = 2; dp[3] = 5; for(int i = 4; i <= N; i++){ dp[i] = add(2 * dp[i − 1], dp[i − 3]); } return dp[N]; } }; main(){ Solution ob; cout << (ob.solve(3)); }

## Input

3

## Output

5