Sum of all minimum occurring elements in an Array
Last Updated :
11 Jul, 2025
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 <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;
}
// 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<Integer,Integer> 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<Integer,Integer> 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 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# 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
<script>
// JavaScript program to find
// the sum of all minimum
// occurring elements in an array
// Function to find the sum of all minimum
// occurring elements in an array
function findSum(arr,N)
{
// Store frequencies of elements
// of the array
let mp = new Map();
for (let i = 0 ; i < N; i++)
{
if(mp.has(arr[i]))
{
mp.set(arr[i], mp.get(arr[i])+1);
}
else
{
mp.set(arr[i], 1);
}
}
// Find the min frequency
let minFreq = Number.MAX_VALUE;
for (let [key, value] of mp.entries())
{
if (value < minFreq)
{
minFreq = value;
}
}
// Traverse the map again and find the sum
let sum = 0;
for (let [key, value] of mp.entries())
{
if (value == minFreq)
{
sum += key * value;
}
}
return sum;
}
// Driver Code
let arr=[ 10, 20, 30, 40, 40 ];
let N = arr.length;
document.write(findSum(arr, N));
// This code is contributed by patel2127
</script>
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N) because it is using unordered_map "mp"
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