# HackerEarth Taxi Please! problem solution

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.

## HackerEarth Taxi Please! problem solution.

`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);    }}`