查找另一个数组中存在的数组元素的频率

给定两个大小分别为N和M的数组arr[]和brr[] ，任务是在arr[]中找到数组brr[]的元素的频率。

例子：

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[]

方法：通过将数组arr[]元素的频率存储在哈希图中，可以轻松解决该任务。遍历数组brr[]并检查它是否存在于 hashmap 中，并存储相应的频率。
下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
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 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 occ=new HashMap();
 
    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 occ = new Dictionary();
 
        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

// 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.

输出

3 4 1 

时间复杂度：O(N)
辅助空间：O(N)