i dont think its possible for unsorted array in O(logn).. Regards, PAYAL GUPTA , NIT-B
On Thu, Feb 9, 2012 at 6:02 PM, atul anand <[email protected]> wrote: > @payal : this will work for only sorted array .... not for unsorted. > > > On Thu, Feb 9, 2012 at 5:50 PM, payal gupta <[email protected]> wrote: > >> http://www.geeksforgeeks.org/archives/4722 >> O(logn) soln... >> >> Regards, >> PAYAL GUPTA, >> NIT-B >> >> >> On Thu, Feb 9, 2012 at 5:33 PM, atul anand <[email protected]>wrote: >> >>> ignore my comment... >>> >>> >>> On Thu, Feb 9, 2012 at 4:15 PM, Ashish Goel <[email protected]> wrote: >>> >>>> log n is impossible. the other solution i thought was of building a >>>> tree where each node contains value and its count. and then building a heap >>>> out of these counts, but this will be overkill. >>>> >>>> the fact that rest of the n/2 elements are not unique is the killer in >>>> the logic otherwise only n/2+1 elements are sufficient >>>> >>>> Best Regards >>>> Ashish Goel >>>> "Think positive and find fuel in failure" >>>> +919985813081 >>>> +919966006652 >>>> >>>> >>>> >>>> On Thu, Feb 9, 2012 at 2:53 PM, atul anand <[email protected]>wrote: >>>> >>>>> i guess can be done using binary indexed tree. >>>>> >>>>> >>>>> On Thu, Feb 9, 2012 at 2:07 PM, Prakhar Jain <[email protected]>wrote: >>>>> >>>>>> But in this post, we don't have prior information about what can be >>>>>> possible majority element. >>>>>> >>>>>> According to my question, we know that either x is the majority >>>>>> element or there is no majority element. >>>>>> Can we use that information to reduce complexity to O(log n)..??? >>>>>> >>>>>> -- >>>>>> You received this message because you are subscribed to the Google >>>>>> Groups "Algorithm Geeks" group. >>>>>> To post to this group, send email to [email protected]. >>>>>> To unsubscribe from this group, send email to >>>>>> [email protected]. >>>>>> 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 [email protected]. >>>>> To unsubscribe from this group, send email to >>>>> [email protected]. >>>>> 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 [email protected]. >>>> To unsubscribe from this group, send email to >>>> [email protected]. >>>> 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 [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> 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 [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> 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 [email protected]. > To unsubscribe from this group, send email to > [email protected]. > 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 [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
