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# C++ program to find out the number of ways a grid with boards can be colored

Suppose, we are given a grid that has 2 rows and n columns. The grid has to be covered by n boards without one board getting over another. Now, the boards have to be colored by any one color between red, blue, and green. Two boards that are adjacent to each other cannot be colored by the same color and if not necessary, all colors do not have to be used. The configuration of the grid is given in the array 'grid', where a particular board in the grid is represented using the same English letter and different boards are represented using different English letters. We have to find out the number of ways the boards can be colored.

So, if the input is like n = 4, grid = {"abbd", "accd"}, then the output will be 6.

There are 6 different ways to color the boards satisfying the given criterion.

## Steps

To solve this, we will follow these steps −

MODVAL := 10^9 + 7 Define an array s for initialize i := 0, when i < n, do: if grid[0, i] is same as grid[1, i], then: insert 1 at the end of s (increase i by 1) Otherwise, insert 2 at the end of s i := i + 2 Define an array tvec if s[0] is same as 1, then: tvec[0] := 3 Otherwise, tvec[0] := 6 for initialize i := 1, when i < size of s, update (increase i by 1), do: if s[i - 1] is same as 2 and s[i] is same as 2, then: tvec[i] := tvec[i - 1] * 3 mod MODVAL if s[i - 1] is same as 2 and s[i] is same as 1, then: tvec[i] := tvec[i - 1] if s[i - 1] is same as 1 and s[i] is same as 2, then: tvec[i] := tvec[i - 1] * 2 mod MODVAL if s[i - 1] is same as 1 and s[i] is same as 1, then: tvec[i] := tvec[i - 1] * 2 mod MODVAL return tvec[size of s - 1]

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h> using namespace std; int solve(int n, vector<string> grid){ int MODVAL = 1e9 + 7; vector<int> s; for (int i = 0; i < n;) { if (grid[0][i] == grid[1][i]) { s.push_back(1); i++; } else { s.push_back(2); i += 2; } } vector<int> tvec(s.size()); if (s[0] == 1) tvec[0] = 3; else tvec[0] = 6; for (int i = 1; i < (int)s.size(); i++) { if (s[i - 1] == 2 && s[i] == 2) tvec[i] = tvec[i - 1] * 3 % MODVAL; if (s[i - 1] == 2 && s[i] == 1) tvec[i] = tvec[i - 1]; if (s[i - 1] == 1 && s[i] == 2) tvec[i] = tvec[i - 1] * 2 % MODVAL; if (s[i - 1] == 1 && s[i] == 1) tvec[i] = tvec[i - 1] * 2 % MODVAL; } return tvec[s.size() - 1]; } int main() { int n = 4; vector <string> grid = {"abbd", "accd"}; cout<< solve(n, grid); return 0; }

## Input

4, {"abbd", "accd"}

## Output

6

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