Related Articles
Sum of all minimum occurring elements in an Array
• Difficulty Level : Basic
• Last Updated : 16 Oct, 2020

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

```Input : arr[] = {1, 1, 2, 2, 3, 3, 3, 3}
Output : 2
The least occurring element is 1 and 2 and it's number
of occurrence is 2. Therefore sum of all 1's and 2's in the
array = 1+1+2+2 = 6.

Input : arr[] = {10, 20, 30, 40, 40}
Output : 60
Elements with least frequency are 10, 20, 30.
Their sum = 10 + 20 + 30 = 60.

```

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 minimum occurring element.
• Now, to find the sum traverse the map again and for all elements with minimum frequency find frequency_of_min_occurring_element*min_occurring_element and find their sum.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the sum of all minimum``// occurring elements in an array` `#include ``using` `namespace` `std;` `// Function to find the sum of all minimum``// 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 min frequency``    ``int` `minFreq = INT_MAX;``    ``for` `(``auto` `itr = mp.begin(); itr != mp.end(); itr++) {``        ``if` `(itr->second < minFreq) {``            ``minFreq = 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 == minFreq) {``            ``sum += itr->first * itr->second;``        ``}``    ``}` `    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 10, 20, 30, 40, 40 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << findSum(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the sum of all minimum``// occurring elements in an array``import` `java.util.Collections;``import` `java.util.Comparator;``import` `java.util.HashMap;``import` `java.util.Iterator;``import` `java.util.Map;` `class` `GFG``{``// Function to find the sum of all minimum``// 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++)``        ``mp.put(arr[i],mp.get(arr[i])==``null``?``1``:mp.get(arr[i])+``1``);`  `    ``// Find the min frequency``    ``int` `minFreq = Integer.MAX_VALUE;``    ``minFreq = Collections.min(mp.entrySet(),``            ``Comparator.comparingInt(Map.Entry::getKey)).getValue();`  `    ``// Traverse the map again and find the sum``    ``int` `sum = ``0``;``    ``for` `(Map.Entry entry : mp.entrySet())``    ``{``        ``if` `(entry.getValue() == minFreq)``        ``{``            ``sum += entry.getKey() * entry.getValue();``        ``}``    ``}` `    ``return` `sum;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``10``, ``20``, ``30``, ``40``, ``40` `};` `    ``int` `N = arr.length;` `    ``System.out.println( findSum(arr, N));``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 program to find theSum of all``# minimum occurring elements in an array``import` `math as mt` `# Function to find theSum of all minimum``# occurring elements in an array``def` `findSum(arr, N):` `    ``# Store frequencies of elements``    ``# of the array``    ``mp ``=` `dict``()``    ``for` `i ``in` `arr:``        ``if` `i ``in` `mp.keys():``            ``mp[i] ``+``=` `1``        ``else``:``            ``mp[i] ``=` `1` `    ``# Find the min frequency``    ``minFreq ``=` `10``*``*``9``    ``for` `itr ``in` `mp:``        ``if` `mp[itr]< minFreq:``            ``minFreq ``=` `mp[itr]``        ` `    ``# Traverse the map again and``    ``# find theSum``    ``Sum` `=` `0``    ``for` `itr ``in` `mp:``        ``if` `mp[itr]``=``=` `minFreq:``            ``Sum` `+``=` `itr ``*` `mp[itr]``        ` `    ``return` `Sum` `# Driver Code``arr ``=` `[ ``10``, ``20``, ``30``, ``40``, ``40``]` `N ``=` `len``(arr)` `print``(findSum(arr, N))` `# This code is contributed by``# mohit kumar 29`

## C#

 `// C# program to find the sum of all minimum``// occurring elements in an array``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the sum of all minimum``// 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]))``        ``{``            ``mp[arr[i]]++;``        ``}``        ``else``        ``{``            ``mp.Add(arr[i], 1);``        ``}``    ``}``  ` `    ``// Find the min frequency``    ``int` `minFreq = Int32.MaxValue;``    ``foreach``(KeyValuePair<``int``, ``int``> itr ``in` `mp)``    ``{``        ``if` `(itr.Value < minFreq)``        ``{``            ``minFreq = itr.Value;``        ``}``    ``}``   ` `    ``// Traverse the map again and find the sum``    ``int` `sum = 0;``    ``foreach``(KeyValuePair<``int``, ``int``> itr ``in` `mp)``    ``{``        ``if` `(itr.Value == minFreq)``        ``{``            ``sum += itr.Key * itr.Value;``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 10, 20, 30, 40, 40 };` `    ``int` `N = arr.Length;``  ` `    ``Console.Write(findSum(arr, N));``}``}` `// This code is contributed by divyeshrabadiya07`
Output:
```60

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up