查找频率在[l，r]范围内的数组元素

给定一个整数数组，请从该数组中查找其频率在[l，r]范围内的元素。

例子：

Input : arr[] = { 1, 2, 3, 3, 2, 2, 5 }
        l = 2, r = 3
Output : 2 3 3 2 2

方法 ：

• 拿一个哈希图，它将存储数组中所有元素的频率。
• 现在，再次遍历。
• 打印其频率在[l，r]范围内的元素。

C++

// C++ program to find the elements whose
// frequency lies in the range [l, r]
#include "iostream"
#include "unordered_map"
using namespace std;
  
void findElements(int arr[], int n, int l, int r)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    unordered_map mp;
  
    for (int i = 0; i < n; ++i) {
  
        // Increment the frequency
        // of the element by 1.
        mp[arr[i]]++;
    }
  
    for (int i = 0; i < n; ++i) {
  
        // Print the element whose frequency
        // lies in the range [l, r]
        if (l <= mp[arr[i]] && mp[arr[i] <= r]) {
            cout << arr[i] << " ";
        }
    }
}
  
int main()
{
    int arr[] = { 1, 2, 3, 3, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int l = 2, r = 3;
    findElements(arr, n, l, r);
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;
  
// Java program to find the elements whose
// frequency lies in the range [l, r]
public class GFG {
  
    static void findElements(int arr[], int n, int l, int r) {
        // Hash map which will store the
        // frequency of the elements of the array.
        Map mp = new HashMap();
  
        for (int i = 0; i < n; ++i) {
  
            // Increment the frequency
            // of the element by 1.
            int a=0;
            if(mp.get(arr[i])==null){
                a=1;
            }
            else{
                a = mp.get(arr[i])+1;
            }
            mp.put(arr[i], a);
        }
  
        for (int i = 0; i < n; ++i) {
  
            // Print the element whose frequency
            // lies in the range [l, r]
            if (l <= mp.get(arr[i]) && (mp.get(arr[i]) <= r)) {
                System.out.print(arr[i] + " ");
            }
        }
    }
  
    public static void main(String[] args) {
        int arr[] = {1, 2, 3, 3, 2, 2, 5};
        int n = arr.length;
        int l = 2, r = 3;
        findElements(arr, n, l, r);
  
    }
}
/*This code is contributed by PrinciRaj1992*/

Python3

# Python 3 program to find the elements whose
# frequency lies in the range [l, r]
  
def findElements(arr, n, l, r):
      
    # Hash map which will store the
    # frequency of the elements of the array.
    mp = {i:0 for i in range(len(arr))}
  
    for i in range(n):
          
        # Increment the frequency
        # of the element by 1.
        mp[arr[i]] += 1
  
    for i in range(n):
          
        # Print the element whose frequency
        # lies in the range [l, r]
        if (l <= mp[arr[i]] and mp[arr[i] <= r]):
            print(arr[i], end = " ")
      
# Driver Code
if __name__ == '__main__':
    arr = [1, 2, 3, 3, 2, 2, 5]
    n = len(arr)
    l = 2
    r = 3
    findElements(arr, n, l, r)
      
# This code is contributed by
# Shashank_Sharma

C#

// C# program to find the elements whose
// frequency lies in the range [l, r]
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    static void findElements(int []arr, int n, int l, int r)
    {
        // Hash map which will store the
        // frequency of the elements of the array.
        Dictionary mp = new Dictionary();
  
        for (int i = 0; i < n; ++i) 
        {
  
            // Increment the frequency
            // of the element by 1.
            int a = 0;
            if(!mp.ContainsKey(arr[i]))
            {
                a = 1;
            }
            else
            {
                a = mp[arr[i]]+1;
            }
              
            if(!mp.ContainsKey(arr[i]))
            {
                mp.Add(arr[i], a); 
            }
            else
            {
                mp.Remove(arr[i]);
                mp.Add(arr[i], a);
            }
        }
  
        for (int i = 0; i < n; ++i)
        {
  
            // Print the element whose frequency
            // lies in the range [l, r]
            if (mp.ContainsKey(arr[i]) && 
                        l <= mp[arr[i]] &&
                        (mp[arr[i]] <= r)) 
            {
                Console.Write(arr[i] + " ");
            }
        }
    }
      
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 2, 3, 3, 2, 2, 5};
        int n = arr.Length;
        int l = 2, r = 3;
        findElements(arr, n, l, r);
  
    }
}
  
// This code has been contributed by 29AjayKumar

输出：

2 3 3 2 2

时间复杂度– O(N)