Find frequencies of elements of an array present in another array

Given two arrays arr[] and brr[] of sizes N and M respectively, the task is to find the frequencies of the elements of array brr[] in arr[]. 

Examples:

Input: N = 8, arr[] = {29, 8, 8, 8, 7, 7, 8, 7}, M = 3, brr[] = {7, 8, 29}
Output: {3, 4, 1}
Explanation: Frequencies of 7, 8 and 29 are 3, 4 and 1 respectively in arr[]
Input: arr[] = N = 6, {4, 5, 6, 5, 5, 3}, M = 3, brr[] = {1, 2, 3}
Output: {0, 0, 1}
Explanation: Frequencies of 1, 2 and 3 are 0, 0 and 1 respectively in arr[]

Approach: The task can easily be solved by storing the frequencies of elements of array arr[] in a hashmap. Iterate over the array brr[] and check if it is present in hashmap or not, and store the corresponding frequency.
Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the frequencies of
// elements of brr[] in array arr[]
void solve(int arr[], int brr[], int N, int M)
{

    // Stores the frequency of elements
    // of array arr[]
    unordered_map<int, int> occ;

    for (int i = 0; i < N; i++)
        occ[arr[i]]++;

    // Iterate over brr[]
    for (int i = 0; i < M; i++) {

        // Check if brr[i] is present in
        // occ or not
        if (occ.find(brr[i]) != occ.end()) {
            cout << occ[brr[i]] << " ";
        }
        else {
            cout << 0 << " ";
        }
    }
}

// Driver Code
int main()
{
    int N = 8;
    int arr[N] = { 29, 8, 8, 8, 7, 7, 8, 7 };
    int M = 3;
    int brr[M] = { 7, 8, 29 };

    solve(arr, brr, N, M);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the frequencies of
// elements of brr[] in array arr[]
static void solve(int arr[], int brr[], int N, int M)
{

    // Stores the frequency of elements
    // of array arr[]
    HashMap<Integer,Integer> occ=new HashMap<Integer,Integer>();

    for (int i = 0; i < N; i++)
    {
        if(occ.containsKey(arr[i])){
            occ.put(arr[i], occ.get(arr[i])+1);
        }
        else{
            occ.put(arr[i], 1);
        }
    }

    // Iterate over brr[]
    for (int i = 0; i < M; i++) {

        // Check if brr[i] is present in
        // occ or not
        if (occ.containsKey(brr[i])) {
            System.out.print(occ.get(brr[i])+ " ");
        }
        else {
            System.out.print(0+ " ");
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 8;
    int arr[] = { 29, 8, 8, 8, 7, 7, 8, 7 };
    int M = 3;
    int brr[] = { 7, 8, 29 };

    solve(arr, brr, N, M);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find the frequencies of
# elements of brr[] in array arr[]
def solve(arr, brr, N, M) :

    # Stores the frequency of elements
    # of array arr[]
    occ = dict.fromkeys(arr,0);

    for i in range(N) :
        occ[arr[i]] += 1;

    # Iterate over brr[]
    for i in range(M) :

        # Check if brr[i] is present in
        # occ or not
        if brr[i] in occ :
            print(occ[brr[i]], end= " ");

        else :
            print(0, end = " ");

# Driver Code
if __name__ == "__main__" :

    N = 8;
    arr = [ 29, 8, 8, 8, 7, 7, 8, 7 ];
    M = 3;
    brr = [ 7, 8, 29 ];

    solve(arr, brr, N, M);

    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

    // Function to find the frequencies of
    // elements of brr[] in array arr[]
    static void solve(int[] arr, int[] brr, int N, int M)
    {

        // Stores the frequency of elements
        // of array arr[]
        Dictionary<int, int> occ = new Dictionary<int, int>();

        for (int i = 0; i < N; i++)
        {
            if (occ.ContainsKey(arr[i]))
            {
                occ[arr[i]] = occ[arr[i]] + 1;
            }
            else
            {
                occ[arr[i]] = 1;
            }
        }

        // Iterate over brr[]
        for (int i = 0; i < M; i++)
        {

            // Check if brr[i] is present in
            // occ or not
            if (occ.ContainsKey(brr[i]))
            {
                Console.Write(occ[brr[i]] + " ");
            }
            else
            {
                Console.Write(0 + " ");
            }
        }
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int N = 8;
        int[] arr = { 29, 8, 8, 8, 7, 7, 8, 7 };
        int M = 3;
        int[] brr = { 7, 8, 29 };

        solve(arr, brr, N, M);
    }
}

// This code is contributed by Saurabh Jaiswal

Javascript

<script>
// Javascript program for the above approach

// Function to find the frequencies of
// elements of brr[] in array arr[]
function solve(arr, brr, N, M) {

  // Stores the frequency of elements
  // of array arr[]
  let occ = new Map();

  for (let i = 0; i < N; i++) {
    if (occ.has(arr[i])) {
      occ.set(arr[i], occ.get(arr[i]) + 1)
    } else {
      occ.set(arr[i], 1)
    }
  }

  // Iterate over brr[]
  for (let i = 0; i < M; i++) {

    // Check if brr[i] is present in
    // occ or not
    if (occ.has(brr[i])) {
      document.write(occ.get(brr[i]) + " ");
    }
    else {
      document.write(0 + " ");
    }
  }
}

// Driver Code
let N = 8;
let arr = [ 29, 8, 8, 8, 7, 7, 8, 7 ];
let M = 3;
let brr = [ 7, 8, 29 ];

solve(arr, brr, N, M);

// This code is contributed by gfgking.
</script>

Output

3 4 1 

Time Complexity: O(N)
Auxiliary Space: O(N)