Question

There are N problems numbered 1..N which you need to complete. You’ve arranged the problems in increasing difficulty order, and the ith problem has estimated difficulty level i. You have also assigned a rating vi to each problem. Problems with similar vi values are similar in nature. On each day, you will choose a subset of the problems and solve them. You’ve decided that each subsequent problem solved on the day should be tougher than the previous problem you solved on that day. Also, to not make it boring, consecutive problems you solve should differ in their vi rating by at least K. What is the least number of days in which you can solve all problems?

Input:
The first line contains the number of test cases T. T test cases follow. Each case contains an integer N and K on the first line, followed by integers v1,…,vn on the second line.

Output:
Output T lines, one for each test case, containing the minimum number of days in which all problems can be solved.

Constraints:
1 <= T <= 100
1 <= N <= 300
1 <= vi <= 1000
1 <= K <= 1000

Sample Input:
2
3 2
5 4 7
5 1
5 3 4 5 6

Sample Output:
2
1

This is one of the challenge from interviewstreet.
Below is my approach
Start from 1st question and find out max possible number of question can be solve and remove these questions from the question list.Now again start from first element of the remainning list and do this till now size of the question list is 0.
I am getting wrong answer from this method so looking for some algo to solve this challenge.

Answer 1

Construct a DAG of problems in the following way. Let p_i and p_j be two different problems. Then we will draw a directed edge from p_i to p_j if and only if p_j can be solved directly after p_i on the same day, consecutively. Namely, the following conditions have to be satisfied:

1. i < j, because you should solve the less difficult problem earlier.
2. |v_i – v_j| >= K (the rating requirement).

Now notice that each subset of problems that is chosen to be solved on some day corresponds to the directed path in that DAG. You choose your first problem, and then you follow the edges step by step, each edge in the path corresponds to the pair of problems that have been solved consecutively on the same day. Also, each problem can be solved only once, so any node in our DAG may appear only in exactly one path. And you have to solve all the problems, so these paths should cover all the DAG.

Now we have the following problem: given a DAG of n nodes, find the minimal number of non-crossing directed paths that cover this DAG completely. This is a well-known problem called Path cover. Generally speaking, it is NP-hard. However, our directed graph is acyclic, and for acyclic graphs it can be solved in polynomial time using reduction to the matching problem. Maximal matching problem, in its turn, can be solved using Hopcroft-Karp algorithm, for example. The exact reduction method is easy and can be read, say, on Wikipedia. For each directed edge (u, v) of the original DAG one should add an undirected edge (a_u, b_v) to the bipartite graph, where {a_i} and {b_i} are two parts of size n.

The number of nodes in each part of the resulting bipartite graph is equal to the number of nodes in the original DAG, n. We know that Hopcroft-Karp algorithm runs in O(n^2.5) in the worst case, and 300^2.5 ≈ 1 558 845. For 100 tests this algorithm should take under a 1 second in total.