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 [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.
