@Ravindra To check the particular number square can be written as sum of two squares or not.
If it has any prime factor x such that x % 4 == 1 then only. Now about time complexity. suppose u have given array is. 10 6 13 9 17 4 18 12 1 5. now u can easily skip the numbers(1, 4, 6, 9, 12, 18); and u have to check only for these numbers(5, 10, 13, 17); so time complexity will reduce to O(n * (number of side which can be largest one for any triplet) ). -- 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.
