Maximum mirrors which can transfer light from bottom to right

A square matrix is given in which each cell represents either a blank or an obstacle. We can place mirrors at blank positionl. All mirrors will be situated at 45 degree, i.e. they can transfer light from bottom to right if no obstacle is there in their path.
In this question we need to count how many such mirrors can be placed in square matrix which can transfer light from bottom to right.
Examples: Output for above example is 2.

In above diagram, mirror at (3, 1) and (5, 5) are able
to send light from bottom to right so total possible
mirror count is 2.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can solve this problem by checking position of such mirrors in matrix, the mirror which can transfer light from bottom to right will not have any obstacle in their path i.e.
if a mirror is there at index (i, j) then
there will be no obstacle at index (k, j) for all k, i < k <= N
there will be no obstacle at index (i, k) for all k, j < k <= N
Keeping above two equations in mind, we can find rightmost obstacle at every row in one iteration of given matrix and we can find bottommost obstacle at every column in another iteration of given matrix. After storing these indices in separate array we can check at each index whether it satisfies no obstacle condition or not and then increase the count accordingly.
Below is implemented solution on above concept which requires O(N^2) time and O(N) extra space.

C++

 // 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 horizontaly (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=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=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

 // 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 horizontaly (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 */

C#

 // 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 horizontaly (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

Output:

2

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