How about using 1 variable o(1) Space.. i.e Scan the array and compare the element... at a[1]=a1, a[2]=b1, a[3]=a2,......
first check(a[i]), say b2 in array of size 8, so Its clear b2 must be placed at 8/2+2 position in an array.. so Space=a[8/2+2] a[8/2+2]=b2, the actual place of b2 now a[6] is stored into space, which is b3, since in every even position B is stored initially we can use this property.. so now a[6] i.e. space which contains b3 should be stored in a[7].. pls try further... ;-) or try making a seires... given array of 10 such values.. intially, a1 is at positin b1 should go to 6 posi b3 goes t 8 posi b4 goes to 9 posi a5 goes to 5 posi.. ......................x posi so can we form a series 6-8-9-5-..... depending upon the size of array???????????????????/ I think Enough for now ;( --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups-beta.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---