Number of cells in a matrix that satisfy the given condition | Java implementation of the approach ; Function to return the number of cells in which mirror can be placed ; Update the row array where row [ i ] [ j ] will store whether the current row i contains all 1 s in the columns starting from j ; Update the column array where col [ i ] [ j ] will store whether the current column j contains all 1 s in the rows starting from i ; To store the required result ; For every cell except the last row and the last column ; 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 ; For the last column ; 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 ; Driver code
import java . util . * ; class GFG { static int N = 3 ; static int numberOfCells ( int mat [ ] [ ] ) { boolean [ ] [ ] row = new boolean [ N ] [ N ] ; boolean [ ] [ ] col = new boolean [ N ] [ N ] ; 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 ; } } } 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 ; } } } int cnt = 0 ; for ( int i = 0 ; i < N - 1 ; i ++ ) { for ( int j = 0 ; j < N - 1 ; j ++ ) { if ( row [ i ] [ j ] && col [ i ] [ j ] ) { cnt ++ ; } } } for ( int i = 0 ; i < N ; i ++ ) { if ( col [ i ] [ N - 1 ] ) cnt ++ ; } for ( int j = 0 ; j < N - 1 ; j ++ ) { if ( row [ N - 1 ] [ j ] ) cnt ++ ; } return cnt ; } public static void main ( String [ ] args ) { int mat [ ] [ ] = { { 0 , 1 , 1 } , { 0 , 1 , 1 } , { 0 , 1 , 1 } } ; System . out . print ( numberOfCells ( mat ) ) ; } }