find kth smallest number in O(logn) time - algorithm

Question

Here is the problem, an unsorted array a[n], and I need to find the kth smallest number in range [i, j], and absolutely 1<=i<=j<=n, k<=j-i+1.
Typically I will use quick-find to do the job, but it is not fast enough if there many query requests with different range [i, j], I hardly to figure out a algorithm to do the query in O(logn) time (preprocessing is allowed).
Any idea is appreciated.
PS
Let me make the problem easier to understand. Any kinds of preprocessing is allowed, but the query needs to be done in O(logn) time. And there will be many (more than 1) queries, like find the 1st in range [3,7], or 3rd in range [10,17], or 11th in range [33, 52].
By range [i, j] I mean in the original array, not sorted or something.
For example, a[5] = {3,1,7,5,9}, query 1st in range [3,4] is 5, 2nd in range [1,3] is 5, 3rd in range [0,2] is 7.

Answer 1

If pre-processing is allowed and not counted towards the time complexity, just use that to construct sub-lists so that you can efficiently find the element you're looking for. As with most optimisations, this trades space for time.
Your pre-processing step is to take your original list of n numbers and create a number of new sublists.
Each of these sublists is a portion of the original, starting with the nth element, extending for m elements and then sorted. So your original list of:
{3, 1, 7, 5, 9}
gives you:
list[0][0] = {3}
list[0][1] = {1, 3}
list[0][2] = {1, 3, 7}
list[0][3] = {1, 3, 5, 7}
list[0][4] = {1, 3, 5, 7, 9}
list[1][0] = {1}
list[1][1] = {1, 7}
list[1][2] = {1, 5, 7}
list[1][3] = {1, 5, 7, 9}
list[2][0] = {7}
list[2][1] = {5, 7}
list[2][2] = {5, 7, 9}
list[3][0] = {5}
list[3][1] = {5,9}
list[4][0] = {9}
This isn't a cheap operation (in time or space) so you may want to maintain a "dirty" flag on the list so you only perform it the first time after you do an modifying operation (insert, delete, change).
In fact, you can use lazy evaluation for even more efficiency. Basically set all sublists to an empty list when you start and whenever you perform a modifying operation. Then, whenever you attempt to access a sublist and it's empty, calculate that sublist (and that one only) before trying to get the kth value out of it.
That ensures sublists are evaluated only when needed and cached to prevent unnecessary recalculation. For example, if you never ask for a value from the 3-through-6 sublist, it's never calculated.
The pseudo-code for creating all the sublists is basically (for loops inclusive at both ends):
for n = 0 to a.lastindex:
create array list[n]
for m = 0 to a.lastindex - n
create array list[n][m]
for i = 0 to m:
list[n][m][i] = a[n+i]
sort list[n][m]
The code for lazy evaluation is a little more complex (but only a little), so I won't provide pseudo-code for that.
Then, in order to find the kth smallest number in the range i through j (where i and j are the original indexes), you simply look up lists[i][j-i][k-1], a very fast O(1) operation:
+--------------------------+
| |
| v
1st in range [3,4] (values 5,9), list[3][4-3=1][1-1-0] = 5
2nd in range [1,3] (values 1,7,5), list[1][3-1=2][2-1=1] = 5
3rd in range [0,2] (values 3,1,7), list[0][2-0=2][3-1=2] = 7
| | ^ ^ ^
| | | | |
| +-------------------------+----+ |
| |
+-------------------------------------------------+
Here's some Python code which shows this in action:
orig = [3,1,7,5,9]
print orig
print "====="
list = []
for n in range (len(orig)):
list.append([])
for m in range (len(orig) - n):
list[-1].append([])
for i in range (m+1):
list[-1][-1].append(orig[n+i])
list[-1][-1] = sorted(list[-1][-1])
print "(%d,%d)=%s"%(n,m,list[-1][-1])
print "====="
# Gives xth smallest in index range y through z inclusive.
x = 1; y = 3; z = 4; print "(%d,%d,%d)=%d"%(x,y,z,list[y][z-y][x-1])
x = 2; y = 1; z = 3; print "(%d,%d,%d)=%d"%(x,y,z,list[y][z-y][x-1])
x = 3; y = 0; z = 2; print "(%d,%d,%d)=%d"%(x,y,z,list[y][z-y][x-1])
print "====="
As expected, the output is:
[3, 1, 7, 5, 9]
=====
(0,0)=[3]
(0,1)=[1, 3]
(0,2)=[1, 3, 7]
(0,3)=[1, 3, 5, 7]
(0,4)=[1, 3, 5, 7, 9]
(1,0)=[1]
(1,1)=[1, 7]
(1,2)=[1, 5, 7]
(1,3)=[1, 5, 7, 9]
(2,0)=[7]
(2,1)=[5, 7]
(2,2)=[5, 7, 9]
(3,0)=[5]
(3,1)=[5, 9]
(4,0)=[9]
=====
(1,3,4)=5
(2,1,3)=5
(3,0,2)=7
=====

Answer 2

Current solution is O( (logn)^2 ). I am pretty sure it can be modified to run on O(logn). The main advantage of this algorithm over paxdiablo's algorithm is space efficiency. This algorithm needs O(nlogn) space, not O(n^2) space.
First, the complexity of finding kth smallest element from two sorted arrays of length m and n is O(logm + logn). Complexity of finding kth smallest element from arrays of lengths a,b,c,d.. is O(loga+logb+.....).
Now, sort the whole array and store it. Sort the first half and second half of the array and store it and so on. You will have 1 sorted array of length n, 2 sorted of arrays of length n/2, 4 sorted arrays of length n/4 and so on. Total memory required = 1*n+2*n/2+4*n/4+8*n/8...= nlogn.
Once you have i and j figure out the list of of subarrays which, when concatenated, give you range [i,j]. There are going to be logn number of arrays. Finding kth smallest number among them would take O( (logn)^2) time.
Example for the last paragraph:
Assume the array is of size 8 (indexed from 0 to 7). You have the following sorted lists:
A:0-7, B:0-3, C:4-7, D:0-1, E:2-3, F:4-5, G:6-7.
Now construct a tree with pointers to these arrays such that every node contains its immediate constituents. A will be root, B and C are its children and so on.
Now implement a recursive function that returns a list of arrays.
def getArrays(node, i, j):
if i==node.min and j==node.max:
return [node];
if i<=node.left.max:
if j<=node.left.max:
return [getArrays(node.left, i, j)]; # (i,j) is located within left node
else:
return [ getArrays(node.left, i, node.left.max), getArrays(node.right, node.right.min, j) ]; # (i,j) is spread over left and right node
else:
return [getArrays(node.right, i, j)]; # (i,j) is located within right node

Answer 3

Preprocess: Make an nxn array where the [k][r] element is the kth smallest element of the first r elements (1-indexed for convenience).
Then, given some particular range [i,j] and value for k, do the following:
Find the element at the [k][j] slot of the matrix; call this x.
go down the i-1 column of your matrix and find how many values in it are smaller than or equal to x (treat column 0 as having 0 smaller entries). By construction, this column will be sorted (all columns will be sorted), so it can be found in log time. Call this value s
Find the element in the [k+s][j] slot of the matrix. This is your answer.
E.g., given 3 1 7 5 9
3 1 1 1 1
X 3 3 3 3
X X 7 5 5
X X X 7 7
X X X X 9
Now, if we're asked for the 2nd smallest in [2,4] range (again, 1-indexing), I first find the 2nd smallest in [1,4] range which is 3. I then look at column 1 and see that there is 1 element less than or equal to 3. Finally, I find the 3rd smallest in [1,4] range at [3][5] slot which is 5, as desired.
This takes n^2 space, and log(n) lookup time.

Answer 4

This one does not require pre-process but is somehow slower than O(logN). It's significantly faster than a naive iterate&count, and could support dynamic modification on the sequence.
It goes like this. Suppose the length n has n=2^x for some x. Construct a segment-tree whose root node represent [0,n-1]. For each of the node, if it represent a node [a,b], b>a, let it has two child nodes each representing [a,(a+b)/2], [(a+b)/2+1,b]. (That is, do a recursive divide-by-two).
Then, on each node, maintain a separate binary search tree for the numbers within that segment. Therefore, each modification on the sequence takes O(logN)[on the segement]*O(logN)[on the BST]. Queries can be done like this, Let Q(a,b,x) be rank of x within segment [a,b]. Obviously, if Q(a,b,x) can be computed efficiently, a binary search on x can compute the answer desired effectively (with an extra O(logE) factor.
Q(a,b,x) can be computed as: find smallest number of segments that make up [a,b], which can be done in O(logN) on the segment tree. For each segment, query on the binary search tree for that segment for the number of elements less than x. Add all these numbers to get Q(a,b,x).
This should be O(logN*logE*logN). Well not exactly what you have asked for though.

Answer 5

In O(log n) time it's not possible to read all of the elements of the array. Since it's not sorted, and there's no other provided information, this is impossible.

Answer 6

There's no way you can do better than O(n) in both worst and average case. You have to look at every single element.