Sum of elements in an array with frequencies greater than or equal to that element

0 0 7 minutes read

Given an array arr[] of N integers. The task is to find the sum of the elements which have frequencies greater than or equal to that element in the array.

Examples:

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

Approach:

Traverse the array and store the frequencies of all the elements in an unordered_map in C++ or equivalent data structure in any other programming language.
Calculate the sum of elements having frequencies greater than or equal to that element.

Below is the implementation of the above approach:

C++

// 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

// 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

# 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#

// 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 

Javascript

<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>

Output

Time complexity: O(n)

Auxiliary Space: O(n)

Method #2:Using Built in python functions:

Approach:

Calculate the frequencies using Counter() function
Calculate the sum of elements having frequencies greater than or equal to that element.

C++

#include <iostream> 
#include <map> 
using namespace std; 
  
int sumOfElements(int arr[], int n){ 
  
    // Map is used to calculate frequency of elements of array 
    map<int, int> m; 
    for(int i = 0; i < n; i++){ 
        if(m.find(arr[i]) != m.end()){ 
            m[arr[i]]++; 
        } else { 
            m[arr[i]] = 1; 
        } 
    } 
  
    int sum = 0; 
  
    // Traverse the map 
    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; 
} 
  
int main(){ 
    int arr[] = {1, 2, 3, 3, 2, 3, 2, 3, 3}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
  
    cout << sumOfElements(arr, n) << endl; 
  
    return 0; 
} 

Java

// Java program for the above approach 
import java.util.HashMap; 
  
class Main { 
  
    // Function to return the sum of elements 
    // in an array having frequency greater 
    // than or equal to that element 
    public static int sumOfElements(int[] arr, int n) 
    { 
  
        // HashMap is used to calculate frequency of 
        // elements of array 
        HashMap<Integer, Integer> m 
            = new HashMap<Integer, Integer>(); 
        for (int i = 0; i < n; i++) { 
            if (m.containsKey(arr[i])) { 
                m.put(arr[i], m.get(arr[i]) + 1); 
            } 
            else { 
                m.put(arr[i], 1); 
            } 
        } 
  
        int sum = 0; 
  
        // Traverse the HashMap 
        for (Integer key : m.keySet()) { 
  
            // Calculate the sum of elements 
            // having frequencies greater than 
            // or equal to the element itself 
            if (m.get(key) >= key) { 
                sum += 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; 
  
        System.out.println(sumOfElements(arr, n)); 
    } 
} 
  
// This code is contributed by phasing17

Python3

# 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 

C#

using System; 
using System.Collections.Generic; 
  
class MainClass  
{ 
  
  // Function to return the sum of elements 
  // in an array having frequency greater 
  // than or equal to that element 
  public static int SumOfElements(int[] arr, int n) 
  { 
  
    // Dictionary is used to calculate frequency of 
    // elements of array 
    Dictionary<int, int> m = new Dictionary<int, int>(); 
    for (int i = 0; i < n; i++) { 
      if (m.ContainsKey(arr[i])) { 
        m[arr[i]]++; 
      } else { 
        m.Add(arr[i], 1); 
      } 
    } 
  
    int sum = 0; 
  
    // Traverse the Dictionary 
    foreach (KeyValuePair<int, int> kvp in m) { 
      // Calculate the sum of elements 
      // having frequencies greater than 
      // or equal to the element itself 
      if (kvp.Value >= kvp.Key) { 
        sum += kvp.Key; 
      } 
    } 
  
    return sum; 
  } 
  
  // Driver code 
  public static void Main() { 
    int[] arr = { 1, 2, 3, 3, 2, 3, 2, 3, 3 }; 
    int n = arr.Length; 
  
    Console.WriteLine(SumOfElements(arr, n)); 
  } 
} 

Javascript

function sumOfElements(arr, n){ 
  
    // Map function is used to calculate frequency of elements of array 
    let m = new Map(); 
    for(let 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 
    for(let [key, value] of m){ 
  
        // Calculate the sum of elements 
        // having frequencies greater than 
        // or equal to the element itself 
        if(value >= key){ 
            sum += key; 
        } 
    } 
  
    return sum; 
} 
  
// Driver code 
let arr = [1, 2, 3, 3, 2, 3, 2, 3, 3]; 
let n = arr.length; 
  
console.log(sumOfElements(arr, n));

Output

Time Complexity: O(n)

Auxiliary Space: O(n)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!