Is is some what similar to what himanshu1 stated but calculations required are less
It can be found using TOURNAMENT APPROACH
minimum element can be found using n-1 comparison
2 nd minimum will compare with all those who lost against minimum so it contains n/2 elements
So second minimum can be found in
Log(n/2-1) comparison
Similarly third smallest can be found using Log(Log(n/4-1)) comparison
Total complexity O(n-1 +Log(n/2-1)+(log(n/2-1)+log(n/4-1)) )
Which is equal of O(n+ klogn)
k<n
C should be correct