# Sum of all minimum occurring elements in an Array

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // CPP 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 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[0]);        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

Output:

60

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

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Improved By : mohit kumar 29, Rajput-Ji

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