text: Counting sets of 1 s and 0 s in a binary matrix | no of columns ; no of rows ; function to calculate the number of non empty sets of cell ; stores the final answer ; traverses row - wise ; traverses column wise ; at the end subtract n * m as no of single sets have been added twice . ; Driver code
code: var m = 3 ; var n = 2 ; function countSets ( a ) { var res = 0 ; for ( i = 0 ; i < n ; i ++ ) { var u = 0 , v = 0 ; for ( j = 0 ; j < m ; j ++ ) { if ( a [ i ] [ j ] == 1 ) u ++ ; else v ++ ; } res += Math . pow ( 2 , u ) - 1 + Math . pow ( 2 , v ) - 1 ; } for ( i = 0 ; i < m ; i ++ ) { var u = 0 , v = 0 ; for ( j = 0 ; j < n ; j ++ ) { if ( a [ j ] [ i ] == 1 ) u ++ ; else v ++ ; } res += Math . pow ( 2 , u ) - 1 + Math . pow ( 2 , v ) - 1 ; } return res - ( n * m ) ; } var a = [ [ 1 , 0 , 1 ] , [ 0 , 1 , 0 ] ] ; document . write ( countSets ( a ) ) ;