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++

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// 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;
}

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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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Improved By : mohit kumar 29