I agree with skript: Number of ways of doing this is n! One in the first row can be placed in n ways. After one in first row has been placed, we can place One in second row in n-1 ways and so on. So total num of ways is n*(n-1)*...*1 = n!
One possible solution to this problem can be coded as follows: Input: integer n Output: different matrices satisfying given criteria Create an array of integers 1 to n. do { // Omega( n! ) For (i =0 to n-1) // Omega(n) print binary representation of (1 << arr[i] ) upto n bits // Omega(n) } while( Next_Permutation( arr, arr+n ) ); TC: Omega ( n! ) * Omega ( n ) * omega ( n ) SC: O(n)...( I dunno [?] whether Next_permutation uses any extra space or what, so i am not considering it... [?] ) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
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