Delete all odd frequency elements from an Array

Given an array arr containing integers of size N, the task is to delete all the elements from the array that have odd frequencies.

Examples:

Input: arr[] = {3, 3, 3, 2, 2, 4, 7, 7}
Output: {2, 2, 7, 7}
Explanation:
Frequency of 3 = 3
Frequency of 2 = 2
Frequency of 4 = 1
Frequency of 7 = 2
Therefore, the elements 3 and 4 have odd frequencies, and hence the are removed.



Input: arr[] = {1, 3, 3, 1, 2, 5, 6, 5}
Output: {1, 3, 3, 1, 5, 5}

Approach:

Below is the implementation of the above approach:

C++

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// C++ program to removes all odd
// frequency elements from an Array
  
#include <bits/stdc++.h>
using namespace std;
  
// Function that removes the
// elements which have odd
// frequencies in the array
void remove(int arr[], int n)
{
    // Create a map to store the
    // frequency of each element
    unordered_map<int, int> m;
    for (int i = 0; i < n; i++) {
        m[arr[i]]++;
    }
  
    // Remove the elements which
    // have odd frequencies
    for (int i = 0; i < n; i++) {
  
        // If the element has
        // odd frequency then skip
        if ((m[arr[i]] & 1))
            continue;
  
        cout << arr[i] << ", ";
    }
}
  
// Driver code
int main()
{
    int arr[]
        = { 3, 3, 3, 2,
            2, 4, 7, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    remove(arr, n);
  
    return 0;
}

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Java

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// Java program to removes all odd
// frequency elements from an Array
import java.util.*;
  
class GFG {
  
    // Function that removes the
    // elements which have odd
    // frequencies in the array
    static void remove(int arr[], int n) {
        // Create a map to store the
        // frequency of each element
        HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            if (mp.containsKey(arr[i])) {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            } else {
                mp.put(arr[i], 1);
            }
        }
  
        // Remove the elements which
        // have odd frequencies
        for (int i = 0; i < n; i++) {
  
            // If the element has
            // odd frequency then skip
            if ((mp.containsKey(arr[i]) && mp.get(arr[i]) % 2 == 1))
                continue;
  
            System.out.print(arr[i] + ", ");
        }
    }
  
    // Driver code
    public static void main(String[] args) {
        int arr[] = { 3, 3, 3, 2, 2, 4, 7, 7 };
        int n = arr.length;
  
        // Function call
        remove(arr, n);
  
    }
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program to removes all odd 
# frequency elements from an Array 
  
# Function that removes the 
# elements which have odd 
# frequencies in the array 
def remove(arr,  n) :
  
    # Create a map to store the
    # frequency of each element
    m = dict.fromkeys(arr,0);
    for i in range(n) :
        m[arr[i]] += 1;
          
    # Remove the elements which
    # have odd frequencies
    for i in range(n) :
  
        # If the element has 
        # odd frequency then skip
        if ((m[arr[i]] & 1)) :
            continue;
              
        print(arr[i],end= ", "); 
  
# Driver code 
if __name__ == "__main__" :
  
    arr    = [ 3, 3, 3, 2
            2, 4, 7, 7 ]; 
    n = len(arr); 
  
    # Function call 
    remove(arr, n); 
  
# This code is contributed by Yash_R

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C#

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// C# program to removes all odd
// frequency elements from an Array
using System;
using System.Collections.Generic;
  
class GFG {
   
    // Function that removes the
    // elements which have odd
    // frequencies in the array
    static void remove(int []arr, int n) {
  
        // Create a map to store the
        // frequency of each element
        Dictionary<int, int> mp = new Dictionary<int, int>();
        for (int i = 0; i < n; i++) {
            if (mp.ContainsKey(arr[i])) {
                mp[arr[i]] =  mp[arr[i]] + 1;
            } else {
                mp.Add(arr[i], 1);
            }
        }
   
        // Remove the elements which
        // have odd frequencies
        for (int i = 0; i < n; i++) {
   
            // If the element has
            // odd frequency then skip
            if ((mp.ContainsKey(arr[i]) && mp[arr[i]] % 2 == 1))
                continue;
   
            Console.Write(arr[i] + ", ");
        }
    }
   
    // Driver code
    public static void Main(String[] args) {
        int []arr = { 3, 3, 3, 2, 2, 4, 7, 7 };
        int n = arr.Length;
   
        // Function call
        remove(arr, n);
    }
}
  
// This code is contributed by 29AjayKumar

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Output:

2, 2, 7, 7,

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Improved By : Rajput-Ji, Yash_R, 29AjayKumar