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

• Difficulty Level : Basic
• Last Updated : 22 Sep, 2022

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++

 `// 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);``    ``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<``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`

## Javascript

 ``

Output

`2 3 3 2 2 `

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

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