I am trying to understand the algorithm that gives me the number of increasing subsequences of length K in an array in time O(nklog(n)). I know how to solve this very same problem using the O(k*n^2) algorithm. I have looked up and found out this solution uses BIT (Fenwick Tree) and DP. I have also found some code, but I have not been able to understand it.
Here are some links I've visited that have been helpful.
Here in SO
Topcoder forum
Random webpage
I would really appreciate if some can help me out understand this algorithm.
I am reproducing my algorithm from here, where its logic is explained:
dp[i, j] = same as before num[i] = how many subsequences that end with i (element, not index this time)
have a certain length
for i = 1 to n do dp[i, 1] = 1
for p = 2 to k do // for each length this time num = {0}
for i = 2 to n do
// note: dp[1, p > 1] = 0
// how many that end with the previous element
// have length p - 1
num[ array[i - 1] ] += dp[i - 1, p - 1] *1*
// append the current element to all those smaller than it
// that end an increasing subsequence of length p - 1,
// creating an increasing subsequence of length p
for j = 1 to array[i] - 1 do *2*
dp[i, p] += num[j]
You can optimize *1*
and *2*
by using segment trees or binary indexed trees. These will be used to efficiently process the following operations on the num
array:
- Given
(x, v)
add v
to num[x]
(relevant for *1*
);
- Given
x
, find the sum num[1] + num[2] + ... + num[x]
(relevant for *2*
).
These are trivial problems for both data structures.
Note: This will have complexity O(n*k*log S)
, where S
is the upper bound on the values in your array. This may or may not be good enough. To make it O(n*k*log n)
, you need to normalize the values of your array prior to running the above algorithm. Normalization means converting all of your array values into values lower than or equal to n
. So this:
5235 223 1000 40 40
Becomes:
4 2 3 1 1
This can be accomplished with a sort (keep the original indexes).