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++
// 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; } |
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<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*/ |
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 |
Output:
2 3 3 2 2
Time Complexity – O(N)
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