Suppose we want to arrange the $n$ numbers stored in any array such that all negative values occur before all positive ones. Minimum number of exchanges required in the worst case is
Answer is (D) None of these
We just require $n/2$ swaps in the worst case. The algorithm is as given below:
Find positive number from left side and negative number from right side and do exchange. Since, at least one of them must be less than or equal to $n/2$, there cannot be more than $n/2$ exchanges. An implementation is given below:
can I use simple selection sort instead ? coz it will too lead to a minimum of n/2 exchange.
If "Rounded to the nearest integer" is the ...