HackerEarth Stevie problem solution

In this HackerEarth Stevie! problem solution You have been given 2 integer arrays A and B each of size N. Now we call a pair of indices (i,j) connected if i = j or A[i] = A[j].

Now, for each index i in the array A where 1 <= i <= N, you need to find the maximum B[j] such that indices i and j are connected. Can you do it ?

HackerEarth Stevie! problem solution.

import java.io.*;
import java.util.*;
import java.math.*;

public final class solution
{
    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 Map<Integer,Integer> m1=new HashMap<>();
  static int max_val=(int)(1e9);
  
  static void put(int idx,int val)
  {
    if(m1.get(idx)==null) 
    {
      m1.put(idx,val);
    }
    
    else
    {
      m1.put(idx,Math.max(m1.get(idx),val));
    }
  }
  
    public static void main(String args[]) throws Exception
    {
    int n=sc.nextInt();int[] a=new int[n],b=new int[n];
    
    if(n<1 || n>200000) throw new Exception("Violation of Constraints");
    
    for(int i=0;i<n;i++)
    {
      a[i]=sc.nextInt();
      
      if(a[i]<1 || a[i]>max_val) throw new Exception("Violation of Constraints");
    }
    
    for(int i=0;i<n;i++)
    {
      b[i]=sc.nextInt();
      
      if(b[i]<1 || b[i]>max_val) throw new Exception("Violation of Constraints");
      
      put(a[i],b[i]);
    }
    
    for(int i=0;i<n;i++)
    {
      out.print(m1.get(a[i])+" ");
    }
    
    out.println("");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());
    }
}

Second solution

#include <bits/stdc++.h>
using namespace std;
int main()
{
  int n;
  cin>>n;
  vector <int> A(n), B(n);
  for (int i = 0; i < n; ++i)
  {
    cin>>A[i];
  }
  for (int i = 0; i < n; ++i)
  {
    cin>>B[i];
  }
  map <int,int> val;
  for (int i = 0; i < n; ++i)
  {
    val[A[i]] = max(B[i], val[A[i]]);
  }
  for (int i = 0; i < n; ++i)
  {
    cout<<val[A[i]]<<" ";
  }
  return 0;
}