View source for TADM2E 4.16

the general form of this problem is to find the kth largest value.  finding the median is when k = n/2.

to find the kth largest value,
* select a partition element and split the array into 2 sub-arrays - one with the elements smaller than the partition and one with the elements larger than the partition. '''O(n)'''
* if the array with the elements larger than the partition has k - 1 elements, the partition is the kth largest element
* if the array with the elements larger than the partition has >= k elements, recurse with the same value of k using the larger elements as the new array. '''O(n/2)''' (average case)
* else the median is in the array with elements smaller than the partition so adjust k to account for the large elements being discarded and recurse using the smaller elements as the new array '''O(n/2)''' (average case)

the overall complexity is O(n) since

O(n) + O(n/2) + O(n/4) + ... approaches O(2n) which is just O(n) since 2 is a constant.