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# C++ Program to find out if there is a pattern in a grid

Suppose, we are given a grid of dimensions n * n. We have to detect if there is a cross pattern in the grid, like below −

#...# .#.#. ..#.. .#.#. #...#

The grid can only contain '#' and '.'. We have to detect the pattern and find out how many such patterns are in the grid. The grid and the dimension are given to us as input.

## Problem Category

Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first, and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can be used. This problem can be solved by some basic logic or a brute-force approach. Follow the following contents to understand the approach better.

So, if the input of our problem is like n = 5, and grid =

#...# .#.#. ..#.. .#.#. #...#,

then the output will be 1.

## Steps

To solve this, we will follow these steps −

count := 0 for initialize i := 1, when i < n - 1, update (increase i by 1), do: for initialize j := 1, when j < n - 1, update (increase j by 1), do: if grid[i, j] is same as '#' and grid[i - 1, j - 1] is same as '#' and grid[i - 1, j + 1] is same as '#' and grid[i + 1, j - 1] is same as '#' and grid[i + 1, j + 1] is same as '#', then: (increase count by 1) print(count)

## Example

Let us see the following implementation to get better understanding −

#include<bits/stdc++.h> using namespace std; void solve(int n, vector<string> grid) { int count = 0; for(int i = 1; i < n - 1; i++){ for(int j = 1; j < n - 1; j++){ if(grid[i][j] == '#' && grid[i - 1][j - 1] == '#' && grid[i - 1][j + 1] == '#' && grid[i + 1][j - 1] == '#' && grid[i + 1][j + 1] == '#') count++; } } cout<< count; } int main() { int n = 5; vector<string> grid = {"#...#", ".#.#.", "..#..", ".#.#.", "#...#"}; solve(n, grid); return 0; }

## Input

5, {"#...#", ".#.#.", "..#..", ".#.#.", "#...#"}

## Output

1

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