Question

This is an interview question I saw online and I am not sure I have correct idea for it.

The problem is here:

Design an algorithm to find the two largest elements in a sequence of n numbers.
Number of comparisons need to be n + O(log n)

I think I might choose quick sort and stop when the two largest elements are find?
But not 100% sure about it. Anyone has idea about it please share

Answer 1

Recursively split the array, find the largest element in each half, then find the largest element that the largest element was ever compared against. That first part requires n compares, the last part requires O(log n). Here is an example:

1 2 5 4 9 7 8 7 5 4 1 0 1 4 2 3
2   5   9   8   5   1   4   3
5       9       5       4
9               5
9

At each step I’m merging adjacent numbers and taking the larger of the two. It takes n compares to get down to the largest number, 9. Then, if we look at every number that 9 was compared against (5, 5, 8, 7), we see that the largest one was 8, which must be the second largest in the array. Since there are O(log n) levels in this, it will take O(log n) compares to do this.