Number of cells in a matrix that satisfy the given condition

Last Updated : 20 Oct, 2022

Given an N * N grid consisting of empty cells (denoted by '1') and obstacles (denoted by '0'), the task is to find the number of empty cells in which a mirror can be placed to view the east-side view of grid from the south-side.
Examples:

Input: mat[][] = {
{1, 1, 1},
{1, 1, 0},
{1, 0, 1}}
Output: 2

On clearly observing the image above, it can be seen that the mirror can be placed in only the cell (0, 0) and the cell (2, 2). If any other cell is chosen then the path of light is obstructed by the obstacles.
Input: mat[][] = {
{0, 1, 1},
{0, 1, 1},
{0, 1, 1}}
Output: 6

Naive approach: It is mentioned that the mirror can only be placed in an empty cell if all the cells on its right and below are empty. The naive approach will be to individually check each cell whether it satisfies the condition or not. This approach will take O(n³) time.
Efficient approach: Create two boolean arrays row[][] and col[][] where row[i][j] will store if all the cells in the i^th row after the j^th column (including it) are 1 else it will store false and col[i][j] will store true if all the cells in the j^th column after the i^th row (including it) are 1 else it will store false. Now, for every cell mat[i][j] if both row[i][j] and col[i][j] are true then the current cell is valid else it is not. Count all such valid cells and print the count in the end.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int N = 3;

// Function to return the number of cells
// in which mirror can be placed
int numberOfCells(int mat[][N])
{

    bool row[N][N] = { { false } };
    bool col[N][N] = { { false } };

    // Update the row array where row[i][j]
    // will store whether the current row i
    // contains all 1s in the columns
    // starting from j
    for (int i = 0; i < N; i++) {
        for (int j = N - 1; j >= 0; j--) {
            if (mat[i][j] == 1) {
                row[i][j] = (j + 1 < N)
                                ? row[i][j + 1]
                                : true;
            }
            else {
                row[i][j] = false;
            }
        }
    }

    // Update the column array where col[i][j]
    // will store whether the current column j
    // contains all 1s in the rows starting from i
    for (int j = 0; j < N; j++) {
        for (int i = N - 1; i >= 0; i--) {
            if (mat[i][j] == 1) {
                col[i][j] = (i + 1 < N)
                                ? col[i + 1][j]
                                : true;
            }
            else {
                col[i][j] = false;
            }
        }
    }

    // To store the required result
    int cnt = 0;

    // For every cell except the last
    // row and the last column
    for (int i = 0; i < N - 1; i++) {
        for (int j = 0; j < N - 1; j++) {

            // If the current cell is not blocked
            // and the light can travel from the
            // next row and the next column
            // then the current cell is valid
            if (row[i][j]
                && col[i][j]) {
                cnt++;
            }
        }
    }

    // For the last column
    for (int i = 0; i < N; i++) {
        if (col[i][N - 1])
            cnt++;
    }

    // For the last row, note that the last column
    // is not taken into consideration as the bottom
    // right element has already been considered
    // in the last column previously
    for (int j = 0; j < N - 1; j++) {
        if (row[N - 1][j])
            cnt++;
    }

    return cnt;
}

// Driver code
int main()
{
    int mat[][N] = { { 0, 1, 1 },
                     { 0, 1, 1 },
                     { 0, 1, 1 } };

    cout << numberOfCells(mat);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{
static int N = 3;

// Function to return the number of cells
// in which mirror can be placed
static int numberOfCells(int mat[][])
{
    boolean [][]row = new boolean[N][N];
    boolean [][]col = new boolean[N][N];

    // Update the row array where row[i][j]
    // will store whether the current row i
    // contains all 1s in the columns
    // starting from j
    for (int i = 0; i < N; i++) 
    {
        for (int j = N - 1; j >= 0; j--) 
        {
            if (mat[i][j] == 1) 
            {
                row[i][j] = (j + 1 < N) ? row[i][j + 1]
                                        : true;
            }
            else
            {
                row[i][j] = false;
            }
        }
    }

    // Update the column array where col[i][j]
    // will store whether the current column j
    // contains all 1s in the rows starting from i
    for (int j = 0; j < N; j++) 
    {
        for (int i = N - 1; i >= 0; i--) 
        {
            if (mat[i][j] == 1)
            {
                col[i][j] = (i + 1 < N) ? col[i + 1][j]
                                        : true;
            }
            else
            {
                col[i][j] = false;
            }
        }
    }

    // To store the required result
    int cnt = 0;

    // For every cell except the last
    // row and the last column
    for (int i = 0; i < N - 1; i++)
    {
        for (int j = 0; j < N - 1; j++)
        {

            // If the current cell is not blocked
            // and the light can travel from the
            // next row and the next column
            // then the current cell is valid
            if (row[i][j] && col[i][j])
            {
                cnt++;
            }
        }
    }

    // For the last column
    for (int i = 0; i < N; i++)
    {
        if (col[i][N - 1])
            cnt++;
    }

    // For the last row, note that the last column
    // is not taken into consideration as the bottom
    // right element has already been considered
    // in the last column previously
    for (int j = 0; j < N - 1; j++) 
    {
        if (row[N - 1][j])
            cnt++;
    }
    return cnt;
}

// Driver code
public static void main(String[] args)
{
    int mat[][] = { { 0, 1, 1 },
                    { 0, 1, 1 },
                    { 0, 1, 1 } };

    System.out.print(numberOfCells(mat));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
N = 3

# Function to return the number of cells
# in which mirror can be placed
def numberOfCells(mat):

    row = [[ False for i in range(N)] 
                   for i in range(N)]
    col = [[ False for i in range(N)] 
                   for i in range(N)]

    # Update the row array where row[i][j]
    # will store whether the current row i
    # contains all 1s in the columns
    # starting from j
    for i in range(N):
        for j in range(N - 1, -1, -1):
            if (mat[i][j] == 1):
                if j + 1 < N:
                    row[i][j] = row[i][j + 1]
                else:
                    row[i][j] = True

            else :
                row[i][j] = False

    # Update the column array where col[i][j]
    # will store whether the current column j
    # contains all 1s in the rows starting from i
    for j in range(N):
        for i in range(N - 1, -1, -1):
            if (mat[i][j] == 1):
                if i + 1 < N:
                    col[i][j] = col[i + 1][j]
                else:
                    col[i][j] = True

            else:
                col[i][j] = False

    # To store the required result
    cnt = 0

    # For every cell except the last
    # row and the last column
    for i in range(N - 1):
        for j in range(N - 1):

            # If the current cell is not blocked
            # and the light can travel from the
            # next row and the next column
            # then the current cell is valid
            if (row[i][j] and col[i][j]):
                cnt += 1

    # For the last column
    for i in range(N):
        if (col[i][N - 1]):
            cnt += 1

    # For the last row, note that the last column
    # is not taken into consideration as the bottom
    # right element has already been considered
    # in the last column previously
    for j in range(N - 1):
        if (row[N - 1][j]):
            cnt += 1

    return cnt

# Driver code
mat = [[0, 1, 1],
       [0, 1, 1],
       [0, 1, 1]]

print(numberOfCells(mat))

# This code is contributed by Mohit Kumar

// C# implementation of the approach
using System;

class GFG
{
static int N = 3;

// Function to return the number of cells
// in which mirror can be placed
static int numberOfCells(int [,]mat)
{
    bool [,]row = new bool[N, N];
    bool [,]col = new bool[N, N];

    // Update the row array where row[i,j]
    // will store whether the current row i
    // contains all 1s in the columns
    // starting from j
    for (int i = 0; i < N; i++) 
    {
        for (int j = N - 1; j >= 0; j--) 
        {
            if (mat[i, j] == 1) 
            {
                row[i, j] = (j + 1 < N) ? row[i, j + 1]
                                        : true;
            }
            else
            {
                row[i, j] = false;
            }
        }
    }

    // Update the column array where col[i,j]
    // will store whether the current column j
    // contains all 1s in the rows starting from i
    for (int j = 0; j < N; j++) 
    {
        for (int i = N - 1; i >= 0; i--) 
        {
            if (mat[i, j] == 1)
            {
                col[i, j] = (i + 1 < N) ? col[i + 1, j]
                                        : true;
            }
            else
            {
                col[i, j] = false;
            }
        }
    }

    // To store the required result
    int cnt = 0;

    // For every cell except the last
    // row and the last column
    for (int i = 0; i < N - 1; i++)
    {
        for (int j = 0; j < N - 1; j++)
        {

            // If the current cell is not blocked
            // and the light can travel from the
            // next row and the next column
            // then the current cell is valid
            if (row[i, j] && col[i, j])
            {
                cnt++;
            }
        }
    }

    // For the last column
    for (int i = 0; i < N; i++)
    {
        if (col[i, N - 1])
            cnt++;
    }

    // For the last row, note that the last column
    // is not taken into consideration as the bottom
    // right element has already been considered
    // in the last column previously
    for (int j = 0; j < N - 1; j++) 
    {
        if (row[N - 1, j])
            cnt++;
    }
    return cnt;
}

// Driver code
public static void Main(String[] args)
{
    int [,]mat = {{ 0, 1, 1 },
                  { 0, 1, 1 },
                  { 0, 1, 1 }};

    Console.Write(numberOfCells(mat));
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>
// Javascript implementation of the approach

let N = 3;
// Function to return the number of cells
// in which mirror can be placed
function numberOfCells(mat)
{
    let row = new Array(N);
    let col = new Array(N);
    
    for(let i=0;i<N;i++)
    {
        row[i]=new Array(N);
        col[i]=new Array(N);
        for(let j=0;j<N;j++)
        {
            row[i][j]=0;
            col[i][j]=0;
        }
    }
  
    // Update the row array where row[i][j]
    // will store whether the current row i
    // contains all 1s in the columns
    // starting from j
    for (let i = 0; i < N; i++) 
    {
        for (let j = N - 1; j >= 0; j--) 
        {
            if (mat[i][j] == 1) 
            {
                row[i][j] = (j + 1 < N) ? row[i][j + 1]
                                        : true;
            }
            else
            {
                row[i][j] = false;
            }
        }
    }
  
    // Update the column array where col[i][j]
    // will store whether the current column j
    // contains all 1s in the rows starting from i
    for (let j = 0; j < N; j++) 
    {
        for (let i = N - 1; i >= 0; i--) 
        {
            if (mat[i][j] == 1)
            {
                col[i][j] = (i + 1 < N) ? col[i + 1][j]
                                        : true;
            }
            else
            {
                col[i][j] = false;
            }
        }
    }
  
    // To store the required result
    let cnt = 0;
  
    // For every cell except the last
    // row and the last column
    for (let i = 0; i < N - 1; i++)
    {
        for (let j = 0; j < N - 1; j++)
        {
  
            // If the current cell is not blocked
            // and the light can travel from the
            // next row and the next column
            // then the current cell is valid
            if (row[i][j] && col[i][j])
            {
                cnt++;
            }
        }
    }
  
    // For the last column
    for (let i = 0; i < N; i++)
    {
        if (col[i][N - 1])
            cnt++;
    }
  
    // For the last row, note that the last column
    // is not taken into consideration as the bottom
    // right element has already been considered
    // in the last column previously
    for (let j = 0; j < N - 1; j++) 
    {
        if (row[N - 1][j])
            cnt++;
    }
    return cnt;
}

// Driver code
let mat = [[0, 1, 1 ],
                    [ 0, 1, 1 ],
                    [ 0, 1, 1 ]];
                    
document.write(numberOfCells(mat));


// This code is contributed by patel2127
</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N²)

AshaRamMeena

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Article Tags :

Number of cells in a matrix that satisfy the given condition

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