Quicksort idea to solve Kth-largest in O(N)

Background knowledge

The formula relation of Kth big and its element position (0-indexed array)

0, 3, 6, 7, 8

5 elements
1st big is pos 4 = 5 - 1
2nd big is pos 3 = 5 - 2
K big is pos [length - K]

Idea of solution

How to find Kth-largest in O(N) time complexity?

• quickSort = (list, pivot) =>
  quickSort(list.filter(smaller than pivot))::
  pivot::
  quickSort(list.filter(greater than pivot))
• Let's say this is 0-indexed array, after 1st round of quickSort, say pivot is at index p
• we know that p already stays at the correct position by when the array is fully sorted
• for this is a 0-indexed array, array[p] is the [array.lenth - p]th big one
• if [array.length - p] is K, we already have the answer
• if [array.length - p] < K, we know that Kth-largest falls in right segment, recursively quickSort right segment
• if [array.length - p] > K, we know that Kth-largest falls in left segment, recursively quickSort left segment

Time complexity

N = list length
each subsequent quicksort will scan half of the original segment
we got N + 1/2N + 1/4N + ... + 1 = 2N - N/2^N (summation formula for geometric sequence)
the upper bound part is 2N, so it is O(N) time complexity