Find array elements with frequencies in range [l , r]

Given an array of integers, find the elements from the array whose frequency lies in the range [l, r].

Examples:

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

Approach :



  • Take a hash map, which will store the frequency of all the elements in the array.
  • Now, traverse once again.
  • Print the elements whose frequency lies between the range [l, r].

C++

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// 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<int, int> 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;
}

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Java














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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<Integer, Integer> mp = new HashMap<Integer, Integer>(); 


  


        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*/



















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Python3














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# 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 



















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C#














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// 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<int, int> mp = new Dictionary<int, int>(); 


  


        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 



















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Output:


2 3 3 2 2





Time Complexity – O(N)


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