i am not sure , but this came to me when i first read it here is the idea:- given : 4 5 7 10 12 13
4 should be there boz it is the least. next is 5 , 5-4 =1 which is less that 4 , so 1 should be there next is 7 , 7-4 = 3 which is less than 4 , so 3 should be there next is 10 , 10-4 = 6 which is greater then 4 , so add previous found elements i.e 1,3,4 add them 1+3+4 = 8 , add 1 to 8 = 9 now check ,can we use this number(i.e 9 ) and previous found elements ( 1,3,4) to form 10 ( 9 +1) : yes -> so 9 should be there next is 12 , 12-4 = 8 ( but now greatest element among 1,3,4,9 is 9) and 8 < 9 ,so skip; next is 13 , 13-4 = 9 same reason for skipping as for number 12 before. On Tue, Dec 13, 2011 at 10:28 PM, top coder <topcode...@gmail.com> wrote: > If pairwise sums of 'n' numbers are given in non-decreasing order > identify the individual numbers. If the sum is corrupted print -1 > Example: > i/p: > 4 > 4 5 7 10 12 13 > > o/p: > 1 3 4 9 > > -- > 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. > > -- 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.
