CODEWITHSUNDEEP
algorithmintervals
Love Paper
Love Paper

Reputation: 471

how to solve this (selecting intervals)

I've given some intervals I = {I(1), I(2), ..., I(m)} for I(i) = [a_i, b_i] (1<=a_i<=b_i<=n). You may suppose that intervals cover each other(sorry i'm poor in english), so there's no intervals such as {[1,5], [3,6]}, {[2,5], [5,7]}. And {[1,1], [2,2], ..., [n,n]} must be included in I.

Let's suppose C(i) = b_i - a_i + 1.

I want to find {I(c_1), I(c_2), ..., I(c_k)} that are non overlapped by each other, and C(c_1) + C(c_2) + ... + C(c_k) = T. (1 <= T <= n).

I could find O(n*T) DP solution using Subset Sum problem, and I think it's NP, but I'm not sure. Can I optimize more than O(n*T)?

Upvotes: 0

Views: 333

Answers (1)

amit
amit

Reputation: 178421

The problem is reduceable from the Subset Sum problem (Given a set of numbers and a target number, find out if there is a subset that sums to this target) with a simple reduction:

Given an instance of subset-sum: S={c_1,c_2,..,c_n},T - create an instance of this problem by creating n non overlapping intervals, interval i, with c_i points (easy to do by ascending order). The same T remains.

Now, the answer to the subset-sum problem is true if and only if there is a subset of intervals that sums to T. It is basically the same problem, since all intervals do not overlap each other by definition of the problem.

From this we can conclude - your problem is NP-Hard.

Moreover, if we could solve the problem better then O(T*n), we could use the same approach to solve the subset sum problem better then O(T*n)1,2.
However, AFAIK, best pseudo polynomial solution to subset sum is O(T*n), so if you have such solution - stick with it.

(1) Converting the problem is O(n)
(2) This claim is true for this specific reduction alone, and NOT for the general case of polynomial reductions.

Upvotes: 1

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