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# Find the sum of all highest occurring elements in an Array

• Difficulty Level : Basic
• Last Updated : 28 May, 2021

Given an array of integers containing duplicate elements. The task is to find the sum of all highest occurring elements in the given array. That is the sum of all such elements whose frequency is maximum in the array.
Examples

```Input : arr[] = {1, 1, 2, 2, 2, 2, 3, 3, 3, 3}
Output : 20
The highest occurring elements are 3 and 2 and their
frequency is 4. Therefore sum of all 3's and 2's in the
array = 3+3+3+3+2+2+2+2 = 20.

Input : arr[] = {10, 20, 30, 40, 40}
Output : 80```

Approach

• Traverse the array and use a unordered_map in C++ to store the frequency of elements of the array such that the key of map is the array element and value is its frequency in the array.
• Then, traverse the map to find the frequency of the max occurring element.
• Now, to find the sum traverse the map again and for all elements with maximum frequency find frequency_of_max_occurring_element*max_occurring_element and find their sum.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the sum of all maximum``// occurring elements in an array` `#include ``using` `namespace` `std;` `// Function to find the sum of all maximum``// occurring elements in an array``int` `findSum(``int` `arr[], ``int` `N)``{``    ``// Store frequencies of elements``    ``// of the array``    ``unordered_map<``int``, ``int``> mp;``    ``for` `(``int` `i = 0; i < N; i++)``        ``mp[arr[i]]++;``    `  `    ``// Find the max frequency``    ``int` `maxFreq = 0;``    ``for` `(``auto` `itr = mp.begin(); itr != mp.end(); itr++) {``        ``if` `(itr->second > maxFreq) {``            ``maxFreq = itr->second;``        ``}``    ``}` `    ``// Traverse the map again and find the sum``    ``int` `sum = 0;``    ``for` `(``auto` `itr = mp.begin(); itr != mp.end(); itr++) {``        ``if` `(itr->second == maxFreq) {``            ``sum += itr->first * itr->second;``        ``}``    ``}` `    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << findSum(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the sum of all maximum``// occurring elements in an array``import` `java.util.*;` `class` `GFG``{` `// Function to find the sum of all maximum``// occurring elements in an array``static` `int` `findSum(``int` `arr[], ``int` `N)``{``    ``// Store frequencies of elements``    ``// of the array``    ``Map mp = ``new` `HashMap<>();``    ``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``);``        ``}``    ``}` `    ``// Find the max frequency``    ``int` `maxFreq = ``0``;``    ``for` `(Map.Entry entry : mp.entrySet())``    ``{``        ``if` `(entry.getValue() > maxFreq)``        ``{``            ``maxFreq = entry.getValue();``        ``}``    ``}` `    ``// Traverse the map again and find the sum``    ``int` `sum = ``0``;``    ``for` `(Map.Entry entry : mp.entrySet())``    ``{``        ``if` `(entry.getValue() == maxFreq)``        ``{``            ``sum += entry.getKey() * entry.getValue();``        ``}``    ``}` `    ``return` `sum;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{` `    ``int` `arr[] = { ``1``, ``1``, ``2``, ``2``, ``2``, ``2``, ``3``, ``3``, ``3``, ``3` `};` `    ``int` `N = arr.length;``    ``System.out.println(findSum(arr, N));``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program to find the Sum of all maximum``# occurring elements in an array` `# Function to find the Sum of all maximum``# occurring elements in an array``def` `findSum(arr, N):``    ` `    ``# Store frequencies of elements``    ``# of the array``    ``mp ``=` `dict``()``    ``for` `i ``in` `range``(N):``        ``mp[arr[i]] ``=` `mp.get(arr[i], ``0``) ``+` `1``    `  `    ``# Find the max frequency``    ``maxFreq ``=` `0``    ``for` `itr ``in` `mp:``        ``if` `(mp[itr] > maxFreq):``            ``maxFreq ``=` `mp[itr]``        `  `    ``# Traverse the map again and find the Sum``    ``Sum` `=` `0``    ``for` `itr ``in` `mp:``        ``if` `(mp[itr] ``=``=` `maxFreq):``            ``Sum` `+``=` `itr``*` `mp[itr]``    ` `    ``return` `Sum`  `# Driver Code` `arr``=` `[``1``, ``1``, ``2``, ``2``, ``2``, ``2``, ``3``, ``3``, ``3``, ``3` `]` `N ``=` `len``(arr)` `print``(findSum(arr, N))` `# This code is contributed by mohit kumar`

## C#

 `// C# program to find the sum of all maximum``// occurring elements in an array``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to find the sum of all maximum``// occurring elements in an array``static` `int` `findSum(``int` `[]arr, ``int` `N)``{``    ``// Store frequencies of elements``    ``// of the array``    ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();``    ``for` `(``int` `i = 0 ; i < N; i++)``    ``{``        ``if``(mp.ContainsKey(arr[i]))``        ``{``            ``var` `val = mp[arr[i]];``            ``mp.Remove(arr[i]);``            ``mp.Add(arr[i], val + 1);``        ``}``        ``else``        ``{``            ``mp.Add(arr[i], 1);``        ``}``    ``}` `    ``// Find the max frequency``    ``int` `maxFreq = 0;``    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp)``    ``{``        ``if` `(entry.Value > maxFreq)``        ``{``            ``maxFreq = entry.Value;``        ``}``    ``}` `    ``// Traverse the map again and find the sum``    ``int` `sum = 0;``    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp)``    ``{``        ``if` `(entry.Value == maxFreq)``        ``{``            ``sum += entry.Key * entry.Value;``        ``}``    ``}` `    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{` `    ``int` `[]arr = { 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 };` `    ``int` `N = arr.Length;``    ``Console.WriteLine(findSum(arr, N));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:

`20`

Time Complexity: O(N), where N is the number of elements in the array.

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