Given an array of integers containing duplicate elements. The task is to find the sum of all highest occurring elements in the given array. That is the sum of all such elements whose frequency is maximum in the array.
Examples:
Input : arr[] = {1, 1, 2, 2, 2, 2, 3, 3, 3, 3} Output : 20 The highest occurring elements are 3 and 2 and their frequency is 4. Therefore sum of all 3's and 2's in the array = 3+3+3+3+2+2+2+2 = 20. Input : arr[] = {10, 20, 30, 40, 40} Output : 80
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 max occurring element.
- Now, to find the sum traverse the map again and for all elements with maximum frequency find frequency_of_max_occurring_element*max_occurring_element and find their sum.
Below is the implementation of the above approach:
C++
// CPP program to find the sum of all maximum // occurring elements in an array #include <bits/stdc++.h> using namespace std;
// Function to find the sum of all maximum // 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 max frequency
int maxFreq = 0;
for ( auto itr = mp.begin(); itr != mp.end(); itr++) {
if (itr->second > maxFreq) {
maxFreq = 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 == maxFreq) {
sum += itr->first * itr->second;
}
}
return sum;
} // Driver Code int main()
{ int arr[] = { 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << findSum(arr, N);
return 0;
} |
Java
// Java program to find the sum of all maximum // occurring elements in an array import java.util.*;
class GFG
{ // Function to find the sum of all maximum // 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++)
{
if (mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i])+ 1 );
}
else
{
mp.put(arr[i], 1 );
}
}
// Find the max frequency
int maxFreq = 0 ;
for (Map.Entry<Integer,Integer> entry : mp.entrySet())
{
if (entry.getValue() > maxFreq)
{
maxFreq = entry.getValue();
}
}
// Traverse the map again and find the sum
int sum = 0 ;
for (Map.Entry<Integer,Integer> entry : mp.entrySet())
{
if (entry.getValue() == maxFreq)
{
sum += entry.getKey() * entry.getValue();
}
}
return sum;
} // Driver Code public static void main(String[] args)
{ int arr[] = { 1 , 1 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 3 };
int N = arr.length;
System.out.println(findSum(arr, N));
} } // This code is contributed by Princi Singh |
Python3
# Python3 program to find the Sum of all maximum # occurring elements in an array # Function to find the Sum of all maximum # occurring elements in an array def findSum(arr, N):
# Store frequencies of elements
# of the array
mp = dict ()
for i in range (N):
mp[arr[i]] = mp.get(arr[i], 0 ) + 1
# Find the max frequency
maxFreq = 0
for itr in mp:
if (mp[itr] > maxFreq):
maxFreq = mp[itr]
# Traverse the map again and find the Sum
Sum = 0
for itr in mp:
if (mp[itr] = = maxFreq):
Sum + = itr * mp[itr]
return Sum
# Driver Code arr = [ 1 , 1 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 3 ]
N = len (arr)
print (findSum(arr, N))
# This code is contributed by mohit kumar |
C#
// C# program to find the sum of all maximum // occurring elements in an array using System;
using System.Collections.Generic;
class GFG
{ // Function to find the sum of all maximum // 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]))
{
var val = mp[arr[i]];
mp.Remove(arr[i]);
mp.Add(arr[i], val + 1);
}
else
{
mp.Add(arr[i], 1);
}
}
// Find the max frequency
int maxFreq = 0;
foreach (KeyValuePair< int , int > entry in mp)
{
if (entry.Value > maxFreq)
{
maxFreq = entry.Value;
}
}
// Traverse the map again and find the sum
int sum = 0;
foreach (KeyValuePair< int , int > entry in mp)
{
if (entry.Value == maxFreq)
{
sum += entry.Key * entry.Value;
}
}
return sum;
} // Driver Code public static void Main(String[] args)
{ int []arr = { 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 };
int N = arr.Length;
Console.WriteLine(findSum(arr, N));
} } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find // the sum of all maximum // occurring elements in an array // Function to find the sum of all maximum // 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 max frequency
let maxFreq = 0;
for (let [key, value] of mp.entries())
{
if (value > maxFreq)
{
maxFreq = value;
}
}
// Traverse the map again and find the sum
let sum = 0;
for (let [key, value] of mp.entries())
{
if (value == maxFreq)
{
sum += key * value;
}
}
return sum;
} // Driver Code let arr=[ 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 ]; let N = arr.length; document.write(findSum(arr, N)); // This code is contributed by patel2127 </script> |
Output:
20
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N) because it is using unordered_map