for this problem, we will have to check all the bits i.e. 32*n where n is the number of integers in the array. n bits in one go. So i think your approach will be fine.
Using bitwise operators will not make any difference. if someone knows better solution ,plz post it. Sanju :) On Sun, Aug 21, 2011 at 12:47 AM, himanshu kansal < himanshukansal...@gmail.com> wrote: > problem: There is an array containing integers..... > for every bit in the integer,you have to print a 1 if no of 1s > corresponding to that bit is more than no of 0s corresponding to that > bit (counting that bit in all the integers) otherwise print a 0(if no > of 0s corresponding to that bit are more). > > this you have to do for all bits in the integers..... > > assumption:integers are of 32bits. > no of integers in array are odd...(i.e. there is no case like no. of > 1s=no. of 0s) > > i have done this by counting the no of 1s and 0s for all bits..... > > but can anyone suggest any other efficient approach (somewhat using > bitwise operators)..... > > -- > 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.
