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 position. 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.

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 <bits/stdc++.h>
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

// 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

# 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#

// 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

Javascript

<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>

Output

2

Time complexity: O(n²).
Auxiliary Space: O(n)