@Sankalp range of input is not given ..so count sort can't taken as an option
On Jan 31, 6:16 pm, sankalp srivastava <richi.sankalp1...@gmail.com> wrote: > Another approach would be to use a counting sort method (less space > efficient though ) > So , for space efficiency , we can use a bitset .element insertion > will take O(n) time in the set (with a space complexity of O(log n) ) > > so , for the above elements , the bitset will look like > (the p# shows the index corresponding to the array) > 0p10p2p3p4p5p6p7 > > Now , start from the 1st element in the bitset , and find the next non- > zero element print it ..continue the loop from this element . > Not very space efficient I suppose , but it is still O(n) and time is > O(n) too . > > @abhijit reddy > Ur simple dp is wrong , you should also process the left part as > pointed by juver++ -- 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.
