In this HackerEarth Taxi Please! problem solution Today, you have been given the task of handling the entire Taxi Network of Berland City. Berland city has a huge number of taxi travellers, and you need to help them in transportation in the most efficient manner.

To be precise, this city consists of N users who want to travel via a Taxi today. You have a total of M taxis and need to cater to the users using these taxis. Each user has two parameters associated with them,  and , denoting the time at which a user requests a taxi and the travel time required to reach the destination of this particular user. Each taxi can be used by a maximum of 1 user at each point in time.

If, at any point in time a user requests a taxi and all M taxis are busy, then this user's request is rejected. If multiple taxis are available at the time of the request of a user, the taxi with the lowest index that is available is alloted to them. Now, you need to find for each user, the index of the taxi alloted to them. If a particular user's request is rejected, print "-1" (without quotes) for them.

`import java.io.*;import java.util.*;class taxi{    static Scanner sc;    static PrintWriter out;    static void init() throws Exception    {        sc=new Scanner(System.in);        out=new PrintWriter(System.out);    }    public static void main(String args[]) throws Exception    {        init();        int n=sc.nextInt(),m=sc.nextInt();        Node[] a=new Node[n];        for(int i=0;i<n;i++)        {            a[i]=new Node(i,sc.nextLong(),sc.nextLong());        }        Arrays.sort(a);        long[] last=new long[m],res=new long[n];        for(int i=0;i<n;i++)        {            boolean used=false;            for(int j=0;j<m;j++)            {                if(last[j]<=a[i].a)                {                    res[a[i].idx]=j+1;                    last[j]=a[i].a+a[i].b;                    used=true;                    break;                }            }            if(!used)            {                res[a[i].idx]=-1;            }        }        for(long x:res)        {            out.print(x+" ");        }        out.close();    }}class Node implements Comparable<Node>{    int idx;    long a,b;    public Node(int idx,long a,long b)    {        this.idx=idx;        this.a=a;        this.b=b;    }    public int compareTo(Node x)    {        return Long.compare(this.a,x.a);    }}`