Suppose sorted file contains following numbers
0,0,0,0 (many times) , 1,1,1,1.... (many times), 2,2,2,2(many times).... 100,100,100,100 (many times)
Now if we want to find top 10 largest numbers then obviously it lies on last 10 indexes of sorted list.
So by sorted list, we can get the answer in O(1) time.
In Min heap, following operations need to be performed
1) Build min heap of size 10 ---> O(10) - constant
2) n-10 comparisons and each time heapify algorithm in the worst case. --
n-10 comparisons + n-10 times heapify algorithms
O(n-10) + O(n-1*log 10) = O(n)+O(n*log 10) = O(n)