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
6
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
6
Time Complexity: O(n)
Auxiliary Space: O(n)
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