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 program to test the above function .
code: m = 3 NEW_LINE n = 2 NEW_LINE def countSets ( a ) : NEW_LINE INDENT res = 0 NEW_LINE for i in range ( n ) : NEW_LINE INDENT u = 0 NEW_LINE v = 0 NEW_LINE for j in range ( m ) : NEW_LINE INDENT if a [ i ] [ j ] : NEW_LINE INDENT u += 1 NEW_LINE DEDENT else : NEW_LINE INDENT v += 1 NEW_LINE DEDENT DEDENT res += pow ( 2 , u ) - 1 + pow ( 2 , v ) - 1 NEW_LINE DEDENT for i in range ( m ) : NEW_LINE INDENT u = 0 NEW_LINE v = 0 NEW_LINE for j in range ( n ) : NEW_LINE INDENT if a [ j ] [ i ] : NEW_LINE INDENT u += 1 NEW_LINE DEDENT else : NEW_LINE INDENT v += 1 NEW_LINE DEDENT DEDENT res += pow ( 2 , u ) - 1 + pow ( 2 , v ) - 1 NEW_LINE DEDENT return res - ( n * m ) NEW_LINE DEDENT a = [ [ 1 , 0 , 1 ] , [ 0 , 1 , 0 ] ] NEW_LINE print ( countSets ( a ) ) NEW_LINE