Suma de elementos en una array con frecuencias mayores o iguales a ese elemento

Dada una array arr[] de N enteros. La tarea es encontrar la suma de los elementos que tienen frecuencias mayores o iguales a ese elemento en la array.
Ejemplos : 

Input: arr[] = {2, 1, 1, 2, 1, 6}
Output: 3
The elements in the array are {2, 1, 6}
Where,
 2 appear 2 times which is greater than equal to 2 itself.
 1 appear 3 times which is greater than 1 itself.
 But 6 appears 1 time which is not greater than or equals to 6.
So, sum = 2 + 1 = 3.

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

Acercarse:  

• Recorra la array y almacene las frecuencias de todos los elementos en un mapa_desordenado en C++ o en una estructura de datos equivalente en cualquier otro lenguaje de programación.
• Calcular la suma de los elementos que tienen frecuencias mayores o iguales a ese elemento.

A continuación se muestra la implementación del enfoque anterior:  

// C++ program to find sum of elements
// in an array having frequency greater
// than or equal to that element
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
int sumOfElements(int arr[], int n)
{
    bool prime[n + 1];
    int i, j;
 
    // Map is used to store
    // element frequencies
    unordered_map<int, int> m;
    for (i = 0; i < n; i++)
        m[arr[i]]++;
 
    int sum = 0;
 
    // Traverse the map using iterators
    for (auto it = m.begin(); it != m.end(); it++) {
 
        // Calculate the sum of elements
        // having frequencies greater than
        // or equal to the element itself
        if ((it->second) >= (it->first)) {
            sum += (it->first);
        }
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << sumOfElements(arr, n);
 
    return 0;
}

// Java program to find sum of elements
// in an array having frequency greater
// than or equal to that element
import java.util.*;
class Solution
{
 
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
static int sumOfElements(int arr[], int n)
{
    boolean prime[] = new boolean[n + 1];
    int i, j;
 
    // Map is used to store
    // element frequencies
    HashMap<Integer, Integer> m= new HashMap<Integer,Integer>();
    for (i = 0; i < n; i++)
        {
            if(m.get(arr[i])==null)
            m.put(arr[i],1);
            else
            m.put(arr[i],m.get(arr[i])+1);
        }
 
    int sum = 0;
        // Getting an iterator
        Iterator hmIterator = m.entrySet().iterator();
   
   
    // Traverse the map using iterators
        while (hmIterator.hasNext()) {
            Map.Entry mapElement = (Map.Entry)hmIterator.next();
 
        // Calculate the sum of elements
        // having frequencies greater than
        // or equal to the element itself
        if (((int)mapElement.getValue()) >= ((int)mapElement.getKey())) {
            sum += ((int)mapElement.getKey());
        }
    }
 
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
    int n = arr.length;
 
    System.out.println(sumOfElements(arr, n));
 
 }
}
//contributed by Arnab Kundu

# Python3 program to find sum of elements
# in an array having frequency greater
# than or equal to that element
 
# Function to return the sum of elements
# in an array having frequency greater
# than or equal to that element
def sumOfElements(arr, n) :
 
    # dictionary is used to store
    # element frequencies
    m = dict.fromkeys(arr, 0)
 
    for i in range(n) :
            m[arr[i]] += 1
 
    sum = 0
 
    # traverse the dictionary
    for key,value in m.items() :
 
        # Calculate the sum of elements
        # having frequencies greater than
        # or equal to the element itself
        if value >= key :
                sum += key
 
    return sum
 
# Driver code
if __name__ == "__main__" :
 
    arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
    n = len(arr)
 
    print(sumOfElements(arr, n))
 
# This code is contributed by Ryuga

// C# program to find sum of elements
// in an array having frequency greater
// than or equal to that element
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return the sum of elements
    // in an array having frequency greater
    // than or equal to that element
    static int sumOfElements(int []arr, int n)
    {
        bool []prime = new bool[n + 1];
        int i;
     
        // Map is used to store
        // element frequencies
        Dictionary<int, int> m= new Dictionary<int,int>();
        for (i = 0; i < n; i++)
            {
                if(!m.ContainsKey(arr[i]))
                    m.Add(arr[i],1);
                else
                {
                    var val = m[arr[i]];
                    m.Remove(arr[i]);
                    m.Add(arr[i], val + 1);
                }
             
            }
     
            int sum = 0;
            // Calculate the sum of elements
            // having frequencies greater than
            // or equal to the element itself
            foreach(KeyValuePair<int, int> entry in m)
            {
                if(entry.Value >= entry.Key)
                {
                    sum+=entry.Key;
                }
            }
     
        return sum;
    }
     
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 1, 2, 3, 3, 2, 3, 2, 3, 3 };
        int n = arr.Length;
     
        Console.WriteLine(sumOfElements(arr, n));
    }
}
 
// This code has been contributed by 29AjayKumar

<script>
// javascript program to find sum of elements
// in an array having frequency greater
// than or equal to that element
 
// Function to return the sum of elements
// in an array having frequency greater
// than or equal to that element
function sumOfElements(arr, n) {
    let prime = new Array(n + 1);
    let i, j;
 
    // Map is used to store
    // element frequencies
    let m = new Map();
    for (i = 0; i < n; i++) {
        if (m.has(arr[i])) {
            m.set(arr[i], m.get(arr[i]) + 1)
        } else[
            m.set(arr[i], 1)
        ]
    }
 
    let sum = 0;
 
    // Traverse the map using iterators
    for (let it of m) {
 
        // Calculate the sum of elements
        // having frequencies greater than
        // or equal to the element itself
        if ((it[1]) >= (it[0])) {
            sum += (it[0]);
        }
    }
 
    return sum;
}
 
// Driver code
let arr = [1, 2, 3, 3, 2, 3, 2, 3, 3];
let n = arr.length;
 
document.write(sumOfElements(arr, n));
 
// This code is contributed by gfgking.
</script>

6

Complejidad temporal: O(n)

Método n.º 2: uso de las funciones integradas de python:

Acercarse:

• Calcule las frecuencias usando la función Counter()
• Calcular la suma de los elementos que tienen frecuencias mayores o iguales a ese elemento.

# Python program for the above approach
from collections import Counter
 
# Function to return the sum of elements
# in an array having frequency greater
# than or equal to that element
def sumOfElements(arr, n):
 
    # Counter function is used to
    # calculate frequency of elements of array
    m = Counter(arr)
 
    sum = 0
 
    # traverse the dictionary
    for key, value in m.items():
 
        # Calculate the sum of elements
        # having frequencies greater than
        # or equal to the element itself
        if value >= key:
            sum += key
 
    return sum
 
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]
    n = len(arr)
 
    print(sumOfElements(arr, n))
 
# This code is contributed by vikkycirus

6

Complejidad de tiempo: O(n)