Suppose, we are given a grid that has 2 rows and n columns. A robot is at position (0, 0) in the grid and wants to visit (1, n - 1) by visiting adjacent and corner cells to its current location. We are given the grid in an array of strings, a cell is blocked if it is marked '#' and it is accessible if it is marked '.'. We have to find out if the robot can visit cell (1, n - 1) from the cell (0, 0).
So, if the input is like n = 4, grid = {".##.", "...."}, then the output will be Possible.
Steps
To solve this, we will follow these steps −
flag := 1
for initialize i := 0, when i < n, update (increase i by 1), do:
if grid[0, i] is same as '#' and grid[1, i] is same as '#',
then:
flag := 0
if flag is same as 0, then:
print("Not Possible.")
Otherwise
print("Possible.")Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
#define N 100
void solve(int n, string grid[]) {
int flag = 1;
for(int i = 0; i < n; i++){
if(grid[0].at(i) == '#' && grid[1].at(i) == '#'){
flag = 0;
}
}
if (flag == 0)
cout<<"Not Possible.";
else
cout<<"Possible.";
}
int main() {
int n = 4;
string grid[] = {".##.", "...."};
solve(n, grid);
return 0;
}Input
4, {".##.", "...."}Output
Possible.