# 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.

**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 <bits/stdc++.h> ` `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[0]); ` ` ` ` ` `cout << findSum(arr, N); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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
[tabbyending]

**Output:**

60

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

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