its coming out be either 1 or 2 in all cases On Sun, Nov 13, 2011 at 1:55 PM, UTKARSH SRIVASTAV <usrivastav...@gmail.com>wrote:
> @Surinder give some proof or logic > > > On Sun, Nov 13, 2011 at 10:25 AM, surender sanke <surend...@gmail.com>wrote: > >> @nitin >> yes i meant the same, if each different character have equal number of >> frequency like abcabcabc a's -3, b's - 3 c's- 3 >> then resultant string size is 2 else 1 >> >> surender >> >> >> On Sun, Nov 13, 2011 at 12:21 AM, Ankur Garg <ankurga...@gmail.com>wrote: >> >>> @Srinivas >>> >>> Wat if the string is abc >>> then it reduces to cc :) ...So size 2 can also be there.so u cant say >>> it will be 1 "always" >>> >>> >>> On Sun, Nov 13, 2011 at 12:01 AM, Srinivasa Chaitanya T < >>> tschaitanya....@gmail.com> wrote: >>> >>>> >>>> >>>> On Sat, Nov 12, 2011 at 4:24 PM, Snoopy Me <thesnoop...@gmail.com>wrote: >>>> >>>>> Given a string consisting of a,b and c's, we can perform the >>>>> following >>>>> operation: >>>>> Take any two adjacent distinct characters and replace it with the >>>>> third character. For example, if 'a' and 'c' are adjacent, they can >>>>> replaced with 'b'. >>>>> What is the smallest string which can result by applying this >>>>> operation repeatedly? >>>>> >>>>> 1) if the string aaaaa......aaa, or bb..bb, etc... string cannot be >>>> modified >>>> 2) if string starts with aaaaac => this can be reduced to aaaab -> aaac >>>> -> aab -> ac -> b >>>> 3) So if string not of type (1), then it can be reduced to single >>>> character "always" using method 2 >>>> e.g: >>>> *aab*cacaab // first reduce aab to b >>>> *bbc*acaab // reduce bbc to c >>>> *ca*caab >>>> *bc*aab >>>> *aaab* >>>> c >>>> .. i guess u got the idea >>>> >>>> >>>> >>>> >>>> >>>>> -- >>>>> 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. >>>>> >>>>> >>>> >>>> >>>> -- >>>> T Srinivasa Chaitanya >>>> >>>> >>>> -- >>>> 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. >>> >> >> -- >> 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. >> > > > > -- > *UTKARSH SRIVASTAV > CSE-3 > B-Tech 3rd Year > @MNNIT ALLAHABAD* > > > -- > 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.
