// C++ program to find how many mirror can transfer // light from bottom to right #include    using namespace std; // method returns number of mirror which can transfer // light from bottom to right int maximumMirrorInMatrix(string mat[] int N) {  // To store first obstacles horizontally (from right)  // and vertically (from bottom)  int horizontal[N] vertical[N];  // initialize both array as -1 signifying no obstacle  memset(horizontal -1 sizeof(horizontal));  memset(vertical -1 sizeof(vertical));  // looping matrix to mark column for obstacles  for (int i=0; i<N; i++)  {  for (int j=N-1; j>=0; j--)  {  if (mat[i][j] == 'B')  continue;  // mark rightmost column with obstacle  horizontal[i] = j;  break;  }  }  // looping matrix to mark rows for obstacles  for (int j=0; j<N; j++)  {  for (int i=N-1; i>=0; i--)  {  if (mat[i][j] == 'B')  continue;  // mark leftmost row with obstacle  vertical[j] = i;  break;  }  }  int res = 0; // Initialize result  // if there is not obstacle on right or below  // then mirror can be placed to transfer light  for (int i = 0; i < N; i++)  {  for (int j = 0; j < N; j++)  {  /* if i > vertical[j] then light can from bottom  if j > horizontal[i] then light can go to right */  if (i > vertical[j] && j > horizontal[i])  {  /* uncomment this code to print actual mirror  position also  cout << i << ' ' << j << endl; */  res++;  }  }  }  return res; } // Driver code to test above method int main() {  int N = 5;  // B - Blank O - Obstacle  string mat[N] = {'BBOBB'  'BBBBO'  'BBBBB'  'BOOBO'  'BBBOB'  };  cout << maximumMirrorInMatrix(mat N) << endl;  return 0; } 

// Java program to find how many mirror can transfer // light from bottom to right import java.util.*; class GFG  {  // method returns number of mirror which can transfer  // light from bottom to right  static int maximumMirrorInMatrix(String mat[] int N)   {  // To store first obstacles horizontally (from right)  // and vertically (from bottom)  int[] horizontal = new int[N];  int[] vertical = new int[N];  // initialize both array as -1 signifying no obstacle  Arrays.fill(horizontal -1);  Arrays.fill(vertical -1);    // looping matrix to mark column for obstacles  for (int i = 0; i < N; i++)   {  for (int j = N - 1; j >= 0; j--)   {  if (mat[i].charAt(j) == 'B')  {  continue;  }  // mark rightmost column with obstacle  horizontal[i] = j;  break;  }  }  // looping matrix to mark rows for obstacles  for (int j = 0; j < N; j++)   {  for (int i = N - 1; i >= 0; i--)   {  if (mat[i].charAt(j) == 'B')   {  continue;  }  // mark leftmost row with obstacle  vertical[j] = i;  break;  }  }  int res = 0; // Initialize result  // if there is not obstacle on right or below  // then mirror can be placed to transfer light  for (int i = 0; i < N; i++)  {  for (int j = 0; j < N; j++)   {  /* if i > vertical[j] then light can from bottom  if j > horizontal[i] then light can go to right */  if (i > vertical[j] && j > horizontal[i])  {  /* uncomment this code to print actual mirror  position also  cout << i << ' ' << j << endl; */  res++;  }  }  }  return res;  } // Driver code public static void main(String[] args)  {  int N = 5;  // B - Blank O - Obstacle  String mat[] = {'BBOBB'  'BBBBO'  'BBBBB'  'BOOBO'  'BBBOB'  };  System.out.println(maximumMirrorInMatrix(mat N)); } } /* This code is contributed by PrinciRaj1992 */ 

# Python3 program to find how many mirror can transfer # light from bottom to right # method returns number of mirror which can transfer # light from bottom to right def maximumMirrorInMatrix(mat N): # To store first obstacles horizontally (from right) # and vertically (from bottom) horizontal = [-1 for i in range(N)] vertical = [-1 for i in range(N)]; # looping matrix to mark column for obstacles for i in range(N): for j in range(N - 1 -1 -1): if (mat[i][j] == 'B'): continue; # mark rightmost column with obstacle horizontal[i] = j; break; # looping matrix to mark rows for obstacles for j in range(N): for i in range(N - 1 -1 -1): if (mat[i][j] == 'B'): continue; # mark leftmost row with obstacle vertical[j] = i; break; res = 0; # Initialize result # if there is not obstacle on right or below # then mirror can be placed to transfer light for i in range(N): for j in range(N):    ''' if i > vertical[j] then light can from bottom  if j > horizontal[i] then light can go to right ''' if (i > vertical[j] and j > horizontal[i]):    ''' uncomment this code to print actual mirror  position also''' res+=1; return res; # Driver code to test above method N = 5; # B - Blank O - Obstacle mat = ['BBOBB' 'BBBBO' 'BBBBB' 'BOOBO' 'BBBOB' ]; print(maximumMirrorInMatrix(mat N)); # This code is contributed by rutvik_56. 

// C# program to find how many mirror can transfer // light from bottom to right using System;   class GFG  {  // method returns number of mirror which can transfer  // light from bottom to right  static int maximumMirrorInMatrix(String []mat int N)   {  // To store first obstacles horizontally (from right)  // and vertically (from bottom)  int[] horizontal = new int[N];  int[] vertical = new int[N];  // initialize both array as -1 signifying no obstacle  for (int i = 0; i < N; i++)   {  horizontal[i]=-1;  vertical[i]=-1;  }    // looping matrix to mark column for obstacles  for (int i = 0; i < N; i++)   {  for (int j = N - 1; j >= 0; j--)   {  if (mat[i][j] == 'B')  {  continue;  }  // mark rightmost column with obstacle  horizontal[i] = j;  break;  }  }  // looping matrix to mark rows for obstacles  for (int j = 0; j < N; j++)   {  for (int i = N - 1; i >= 0; i--)   {  if (mat[i][j] == 'B')   {  continue;  }  // mark leftmost row with obstacle  vertical[j] = i;  break;  }  }  int res = 0; // Initialize result  // if there is not obstacle on right or below  // then mirror can be placed to transfer light  for (int i = 0; i < N; i++)  {  for (int j = 0; j < N; j++)   {  /* if i > vertical[j] then light can from bottom  if j > horizontal[i] then light can go to right */  if (i > vertical[j] && j > horizontal[i])  {  /* uncomment this code to print actual mirror  position also  cout << i << ' ' << j << endl; */  res++;  }  }  }  return res;  } // Driver code public static void Main(String[] args)  {  int N = 5;  // B - Blank O - Obstacle  String []mat = {'BBOBB'  'BBBBO'  'BBBBB'  'BOOBO'  'BBBOB'  };  Console.WriteLine(maximumMirrorInMatrix(mat N)); } } // This code is contributed by Princi Singh 

<script> // JavaScript program to find how many mirror can transfer // light from bottom to right // method returns number of mirror which can transfer // light from bottom to right function maximumMirrorInMatrix(mat N)  {  // To store first obstacles horizontally (from right)  // and vertically (from bottom)  var horizontal = Array(N).fill(-1);  var vertical = Array(N).fill(-1);    // looping matrix to mark column for obstacles  for (var i = 0; i < N; i++)   {  for (var j = N - 1; j >= 0; j--)   {  if (mat[i][j] == 'B')  {  continue;  }  // mark rightmost column with obstacle  horizontal[i] = j;  break;  }  }  // looping matrix to mark rows for obstacles  for (var j = 0; j < N; j++)   {  for (var i = N - 1; i >= 0; i--)   {  if (mat[i][j] == 'B')   {  continue;  }  // mark leftmost row with obstacle  vertical[j] = i;  break;  }  }  var res = 0; // Initialize result  // if there is not obstacle on right or below  // then mirror can be placed to transfer light  for (var i = 0; i < N; i++)  {  for (var j = 0; j < N; j++)   {  /* if i > vertical[j] then light can from bottom  if j > horizontal[i] then light can go to right */  if (i > vertical[j] && j > horizontal[i])  {  /* uncomment this code to print actual mirror  position also  cout << i << ' ' << j << endl; */  res++;  }  }  }  return res; } // Driver code var N = 5; // B - Blank O - Obstacle var mat = ['BBOBB'  'BBBBO'  'BBBBB'  'BOOBO'  'BBBOB' ]; document.write(maximumMirrorInMatrix(mat N)); </script>