Print elements of an array according to the order defined by another array | set 2

Given two arrays a1[] and a2[], print elements of a1 in such a way that the relative order among the elements will be the same as those are in a2. That is, elements which come before in the array a2[], print those elements first from the array a1[]. For the elements not present in a2, print them at last in sorted order. 
It is also given that the number of elements in a2[] is smaller than or equal to the number of elements in a1[], and a2[] has all distinct elements.
Example: 
 

Input: a1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8} 
a2[] = {2, 1, 8, 3} 
Output: 2 2 1 1 8 8 3 5 6 7 9
Input: a1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8} 
a2[] = {1, 10, 11} 
Output: 1 1 2 2 3 5 6 7 8 8 9 
 

Simple Approach: We create a temporary array and a visited array where the temporary array is used to copy contents of a1[] to it and visited array is used to mark those elements in the temporary array which are copied to a1[]. Then sorting the temporary array and doing a binary search for every element of a2[] in a1[]. You can find the solution here.
Efficient Approach: We can print the elements of a1[] according to the order defined by a2[] using map in c++ in O(mlog(n)) time. We traverse through a1[] and store the frequency of every number in a map. Then we traverse through a2[] and check if the number is present in the map. If the number is present, then print it that many times and erase the number from the map. Print the rest of the numbers present in the map sequentially as numbers are stored in the map in sorted order.
Below is the implementation of the above approach: 
 

// A C++ program to print an array according
// to the order defined by another array
#include <bits/stdc++.h>
using namespace std;

// Function to print an array according
// to the order defined by another array
void print_in_order(int a1[], int a2[], int n, int m)
{
    // Declaring map and iterator
    map<int, int> mp;
    map<int, int>::iterator itr;

    // Store the frequency of each
    // number of a1[] int the map
    for (int i = 0; i < n; i++)
        mp[a1[i]]++;

    // Traverse through a2[]
    for (int i = 0; i < m; i++) {
        // Check whether number
        // is present in map or not

        if (mp.find(a2[i]) != mp.end()) {
            itr = mp.find(a2[i]);

            // Print that number that
            // many times of its frequency
            for (int j = 0; j < itr->second; j++)
                cout << itr->first << " ";
            mp.erase(a2[i]);
        }
    }

    // Print those numbers that are not
    // present in a2[]
    for (itr = mp.begin(); itr != mp.end(); itr++) {
        for (int j = 0; j < itr->second; j++)
            cout << itr->first << " ";
    }

    cout << endl;
}

// Driver code
int main()
{
    int a1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 };
    int a2[] = { 2, 1, 8, 3 };
    int n = sizeof(a1) / sizeof(a1[0]);
    int m = sizeof(a2) / sizeof(a2[0]);

    print_in_order(a1, a2, n, m);

    return 0;
}

/*package whatever //do not write package name here */
import java.util.*;

class GFG {

  // Function to print an array according
  // to the order defined by another array
  static void print_in_order(int a1[], int a2[], int n, int m)
  {
    // Declaring map and iterator
    HashMap<Integer, Integer> mp = new HashMap<>();

    // Store the frequency of each
    // number of a1[] int the map
    for (int i = 0; i < n; i++)
      mp.put(a1[i],mp.getOrDefault(a1[i],0)+1);  

    // Traverse through a2[]
    for (int i = 0; i < m; i++) {
      // Check whether number
      // is present in map or not

      if (mp.containsKey(a2[i])) {
        // Print that number that
        // many times of its frequency
        for (int j = 0; j < mp.get(a2[i]); j++)
          System.out.print(a2[i] + " ");

        mp.remove(a2[i]);
      }
    }

    // Print those numbers that are not
    // present in a2[]
    for (int i:mp.keySet()) {
      for (int j = 0; j < mp.get(i); j++)
        System.out.print(i + " ");
    }

    System.out.println();
  }

  public static void main (String[] args) {
    int a1[] = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 };
    int a2[] = { 2, 1, 8, 3 };
    int n = a1.length;
    int m = a2.length;

    print_in_order(a1, a2, n, m);

  }
}

// This code is contributed by aadityaburujwale.

# A Python3 program to print an array according 
# to the order defined by another array 

# Function to print an array according 
# to the order defined by another array 
def print_in_order(a1, a2, n, m) : 

    # Declaring map and iterator 
    mp = dict.fromkeys(a1,0); 

    # Store the frequency of each 
    # number of a1[] int the map 
    for i in range(n) :
        mp[a1[i]] += 1; 

    # Traverse through a2[] 
    for i in range(m) :
        # Check whether number 
        # is present in map or not 

        if a2[i] in mp.keys() :

            # Print that number that 
            # many times of its frequency 
            for j in range(mp[a2[i]]) : 
                print(a2[i],end=" "); 
                
            del(mp[a2[i]]); 

    # Print those numbers that are not 
    # present in a2[] 
    for key,value in mp.items() : 
        for j in range(value) :
            print(key,end=" "); 

    print(); 


# Driver code 
if __name__ == "__main__" : 

    a1 = [ 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 ]; 
    a2 = [ 2, 1, 8, 3 ]; 
    n =len(a1); 
    m = len(a2); 

    print_in_order(a1, a2, n, m); 
    
    # This code is contributed by AnkitRai01

using System;
using System.Collections.Generic;

class GFG {
    static void PrintInOrder(int[] a1, int[] a2)
    {
        Dictionary<int, int> dict
            = new Dictionary<int, int>();

        // Store the frequency of each
        // number of a1[] int the dictionary
        for (int i = 0; i < a1.Length; i++) {
            if (dict.ContainsKey(a1[i]))
                dict[a1[i]]++;
            else
                dict.Add(a1[i], 1);
        }

        // Traverse through a2[]
        for (int i = 0; i < a2.Length; i++) {
            // Check whether number
            // is present in dictionary or not
            if (dict.ContainsKey(a2[i])) {
                // Print that number that
                // many times of its frequency
                for (int j = 0; j < dict[a2[i]]; j++)
                    Console.Write(a2[i] + " ");

                dict.Remove(a2[i]);
            }
        }

        // Print those numbers that are not
        // present in a2[]
        foreach(var pair in dict)
        {
            for (int j = 0; j < pair.Value; j++)
                Console.Write(pair.Key + " ");
        }

        Console.WriteLine();
    }

    static void Main(string[] args)
    {
        int[] a1 = { 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 };
        int[] a2 = { 2, 1, 8, 3 };

        PrintInOrder(a1, a2);
    }
}

<script>

// A JavaScript program to print an array according
// to the order defined by another array

// Function to print an array according
// to the order defined by another array
function print_in_order(a1, a2, n, m){

    // Declaring map and iterator
    let mp = new Map();

    // Store the frequency of each
    // number of a1[] int the map
    for(let i=0;i<n;i++){
        if(mp.has(a1[i]))
            mp.set(a1[i],mp.get(a1[i])+1);
        else
            mp.set(a1[i],1);
    }

    // Traverse through a2[]
    for(let i=0;i<m;i++){
        // Check whether number
        // is present in map or not

        if(mp.has(a2[i])){

            // Print that number that
            // many times of its frequency
            for(let j=0;j<mp.get(a2[i]);j++)
                document.write(a2[i]," ");
                
            mp.delete(a2[i]);
        }
    }

    // Print those numbers that are not
    // present in a2[]
    for(let [key,value] of mp)
        for(let j=0;j<value;j++)
            document.write(key," ");

    document.write("</br>");
}

// Driver code
let a1 = [ 2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8 ];
let a2 = [ 2, 1, 8, 3 ];
let n = a1.length;
let m = a2.length;

print_in_order(a1, a2, n, m);

// This code is contributed by shinjanpatra

</script>

2 2 1 1 8 8 3 5 6 7 9