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++
// 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
// 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
# 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 |
Javascript
<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> |
Output:
60
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”