1. Medians and Order Statistics

2. The i th order statistic of a set of n elements is the i th smallest element.
For example, the minimum of a set of elements is the first order statistic (i == 1), and
the maximum is the n-th order statistic (i == n).
A median, informally, is the “halfway point” of the set.

3. We formally specify the selection problem as follows:
Solution:
We can solve the selection problem in O(n lg n) time,
Sort the numbers using heapsort or merge sort and then simply index the i th element in the output array.
We can develop faster algorithms.

4. Finding Minimum and maximum value
We can find the minimum with O(n) comparisons as well.
This is the optimal algorithm.

5. Simultaneous minimum and maximum
Find both the minimum and the maximum of a set of n elements.
We can find the minimum and maximum independently, using n-1 comparisons for each, for a total of 2n -2 comparisons, The complexity is Θ(n).
Compare pairs of elements from the input first with each other, and then compare the smaller with the current minimum and the larger to the current maximum, at a cost of 3 comparisons for every 2 elements

6. Selection in expected linear time
The following code for RANDOMIZED-SELECT returns the i th smallest element of the array A[p,r].
RANDOMIZED-SELECT(A,p,r,i)
if p=r then return A[p] // stopping condition
qRANDOMIZED-PARTITION(A,p,r)
//the q holds for A[p,q-1]A[q] A[q+1,r]
k q-p+1
if i=k then return A[q]
else if i<k
then return RANDOMIZED-SELECT(A,p,q-1,i)
else return RANDOMIZED-SELECT(A,q+1,r,i-k)

8. Selection in worst-case linear time
The worst-case running time for RANDOMIZED-SELECT is
The algorithm has a linear expected running time.
A selection algorithm whose running time is O(n) in the worst case.
We guarantee a good split upon partitioning the array.

9. Guaranteeing a Good Split
Does RANDOMIZED-SELECT guarantee a good split?
Idea: recursively find the median of medians
Divide elements into groups of 5
Find the medians of those five elements
Find the median of those medians
Partition around that median
Your partition point will have k elements to the left, and n -k elements to the right. Make decision.

10. median-of-medians x is labeled.

11. Interesting Results
At least half of the medians are greater than the median of medians
This means 3 of the 5 of those are greater than x is at least
This also means we only have 7n/10 elements left!

12. The Recurrence Equation
T (n) = T (n/5) + T (7n/10 + 6) + O(n)
This turns out to be linear
Time to find the
median of medians
The remaining part
of the problem
Dividing, finding the medians
and partitioning around the
median of medians.

13. Summary
Finding the i-th order statistic using sorting takes (n lg n).
Not necessary to sort everything
Can leverage off of the PARTITION method from QUICKSORT
You can guarantee a good split by finding the median of medians