Given an array of integers, remove 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 : 1 5
We remove all those elements with frequencies from 2 to 5.Input : arr[] = { 1, 2, 3, 3, 3, 3 }
l = 2, r = 3
Output : 3 3 3 3
Approach :
- Take a hash map, which will store the frequency of all the elements in the array.
- Now, traverse once again.
- Remove the elements whose frequency lies between [l, r].
- Else, print it.
Implementation:
C++
// C++ program to remove the elements whose// frequency appears in the range [l, r]#include "iostream"#include "unordered_map"using namespace std;void removeElements(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) { // Incrementing the frequency // of the element by 1. mp[arr[i]]++; } for (int i = 0; i < n; ++i) { // Print the element whose frequency // is not in the range [l, r] if (mp[arr[i]] < l or 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; removeElements(arr, n, l, r); return 0;} |
Java
// Java program to remove the elements whose// frequency appears in the range [l, r]import java.util.HashMap;class GFG { static void removeElements(int arr[], int n, int l, int r) { // Hash map which will store the // frequency of the elements of the array. HashMap<Integer, Integer> mp = new HashMap<>(); for (int i = 0; i < n; ++i) { // Incrementing the frequency // of the element by 1. // mp[arr[i]]++; int val = 0; if (mp.get(arr[i]) == null) { val = 1; } else { val = mp.get(arr[i]) + 1; } // System.out.println("--"+mp.get(arr[i])); mp.put(arr[i], val); } for (int i = 0; i < n; ++i) { // Print the element whose frequency // is not in the range [l, r] if (mp.get(arr[i]) < l || mp.get(arr[i]) > r) { System.out.print(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; removeElements(arr, n, l, r); }}// This code is contributed by 29AjayKumar |
Python3
# Python 3 program to remove the elements # whose frequency appears in the range [l, r]def removeElements(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): # Incrementing the frequency # of the element by 1. mp[arr[i]] += 1 for i in range(n): # Print the element whose frequency # is not in the range [l, r] if (mp[arr[i]] < l or mp[arr[i] > r]): print(arr[i], end = " ") # Driver Codeif __name__ == '__main__': arr = [1, 2, 3, 3, 2, 2, 5] n = len(arr) l = 2 r = 3 removeElements(arr, n, l, r); # This code is contributed by# Sahil_Shelangia |
C#
// C# program to remove the elements whose// frequency appears in the range [l, r]using System;using System.Collections.Generic;class GFG { static void removeElements(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) { // Incrementing the frequency // of the element by 1. // mp[arr[i]]++; int val = 0; if (!mp.ContainsKey(arr[i])) { val = 1; } else { val = mp[arr[i]] + 1; } if (!mp.ContainsKey(arr[i])) mp.Add(arr[i], val); else { mp.Remove(arr[i]); mp.Add(arr[i], val); } } for (int i = 0; i < n; ++i) { // Print the element whose frequency // is not in the range [l, r] if (mp[arr[i]] < l || 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; removeElements(arr, n, l, r); }}// This code contributed by Rajput-Ji |
Javascript
<script>// JavaScript program to remove the elements whose// frequency appears in the range [l, r]function removeElements(arr, n, l, r){ // Hash map which will store the // frequency of the elements of the array. let mp = new Map(); for(let i = 0; i < n; ++i) { // Incrementing the frequency // of the element by 1. // mp[arr[i]]++; let val = 0; if (mp.get(arr[i]) == null) { val = 1; } else { val = mp.get(arr[i]) + 1; } mp.set(arr[i], val); } for(let i = 0; i < n; ++i) { // Print the element whose frequency // is not in the range [l, r] if (mp.get(arr[i]) < l || mp.get(arr[i]) > r) { document.write(arr[i] + " "); } }}// Driver Codelet arr = [ 1, 2, 3, 3, 2, 2, 5 ];let n = arr.length;let l = 2, r = 3;removeElements(arr, n, l, r);// This code is contributed by code_hunt</script> |
Output
1 5
Time Complexity: O(N)
Auxiliary Space: O(N)
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