# HackerEarth Left or Right problem solution

In this HackerEarth Left or Right problem solution, A country X consists of N cities numbered 0 through N - 1. The map of this country can be represented as a cycle — for each valid i, city i has exactly two neighboring cities (i + 1)%N and (i - 1 + N)%N.

The cities in the country are broadly categorized into different types. For each valid i, the type of city i is denoted by Ai.

A person called Suarez wants to travel between some pairs of cities. You are given Q queries. In each query, Suarez wants to travel from city number Y to a city of type Z. Also, he wants to travel only in a given direction along the cycle.

The direction is represented by a letter — L or R. If it's L, Suarez may only move from city i to city (i - 1 + N)%N for each valid i. If it's R, he may only move from city i to city (i + 1)%N.

You need to find the minimum number of steps Suarez needs to take to reach a city of type Z if he starts moving from city Y in the given direction (he can take zero steps). In one step, Suarez can move from his current city to a neighboring city.

If Suarez can never reach a city of type Z for a given query, print 1 instead.

## HackerEarth Left or Right problem solution.

`import java.io.*;import java.util.*;import java.math.*;import java.util.concurrent.*;public final class new_sol{    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));    static FastScanner sc=new FastScanner(br);    static PrintWriter out=new PrintWriter(System.out);    static Random rnd=new Random();    static List<Integer> list=new ArrayList<>();    static int maxn=(int)(2e5+5);        public static void main(String args[]) throws Exception    {        int n=sc.nextInt(),q=sc.nextInt();int[] a=new int[n];                for(int i=0;i<n;i++)        {            a[i]=sc.nextInt();                        list.add(a[i]);        }                Collections.sort(list);int[] rank=new int[maxn];                for(int i=0;i<maxn;i++)        {            rank[i]=Collections.binarySearch(list,i);        }                int[][] l_dis=new int[n+1][n+1],r_dis=new int[n+1][n+1];                for(int i=0;i<=n;i++)        {            for(int j=0;j<=n;j++)            {                l_dis[i][j]=r_dis[i][j]=Integer.MAX_VALUE;            }        }                for(int i=0;i<n;i++)        {            int pos=i,steps=0;                        while(steps<n)            {                int now=rank[a[pos]];                                r_dis[i][now]=Math.min(r_dis[i][now],steps);                                steps++;pos=(pos+1)%n;            }                        pos=i;steps=0;                        while(steps<n)            {                int now=rank[a[pos]];                                l_dis[i][now]=Math.min(l_dis[i][now],steps);                                steps++;pos=(pos-1+n)%n;            }        }                while(q>0)        {                        int idx=sc.nextInt(),type=rank[sc.nextInt()];char ch=sc.next().charAt(0);                        int ans=-1;                        if(ch=='L' && type>=0)            {                ans=l_dis[idx][type];            }                        else if(ch=='R' && type>=0)            {                ans=r_dis[idx][type];            }                        out.println(ans<Integer.MAX_VALUE?ans:-1);                        q--;        }                out.close();    }}class FastScanner{    BufferedReader in;    StringTokenizer st;    public FastScanner(BufferedReader in) {        this.in = in;    }        public String nextToken() throws Exception {        while (st == null || !st.hasMoreTokens()) {            st = new StringTokenizer(in.readLine());        }        return st.nextToken();    }        public String next() throws Exception {        return nextToken().toString();    }        public int nextInt() throws Exception {        return Integer.parseInt(nextToken());    }    public long nextLong() throws Exception {        return Long.parseLong(nextToken());    }    public double nextDouble() throws Exception {        return Double.parseDouble(nextToken());    }}`