Find the sum of all highest occurring elements in an Array
Last Updated :
05 Sep, 2022
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++
#include <bits/stdc++.h>
using namespace std;
int findSum( int arr[], int N)
{
unordered_map< int , int > mp;
for ( int i = 0; i < N; i++)
mp[arr[i]]++;
int maxFreq = 0;
for ( auto itr = mp.begin(); itr != mp.end(); itr++) {
if (itr->second > maxFreq) {
maxFreq = itr->second;
}
}
int sum = 0;
for ( auto itr = mp.begin(); itr != mp.end(); itr++) {
if (itr->second == maxFreq) {
sum += itr->first * itr->second;
}
}
return sum;
}
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
import java.util.*;
class GFG
{
static int findSum( int arr[], int N)
{
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 );
}
}
int maxFreq = 0 ;
for (Map.Entry<Integer,Integer> entry : mp.entrySet())
{
if (entry.getValue() > maxFreq)
{
maxFreq = entry.getValue();
}
}
int sum = 0 ;
for (Map.Entry<Integer,Integer> entry : mp.entrySet())
{
if (entry.getValue() == maxFreq)
{
sum += entry.getKey() * entry.getValue();
}
}
return sum;
}
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));
}
}
|
Python3
def findSum(arr, N):
mp = dict ()
for i in range (N):
mp[arr[i]] = mp.get(arr[i], 0 ) + 1
maxFreq = 0
for itr in mp:
if (mp[itr] > maxFreq):
maxFreq = mp[itr]
Sum = 0
for itr in mp:
if (mp[itr] = = maxFreq):
Sum + = itr * mp[itr]
return Sum
arr = [ 1 , 1 , 2 , 2 , 2 , 2 , 3 , 3 , 3 , 3 ]
N = len (arr)
print (findSum(arr, N))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int findSum( int []arr, int N)
{
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);
}
}
int maxFreq = 0;
foreach (KeyValuePair< int , int > entry in mp)
{
if (entry.Value > maxFreq)
{
maxFreq = entry.Value;
}
}
int sum = 0;
foreach (KeyValuePair< int , int > entry in mp)
{
if (entry.Value == maxFreq)
{
sum += entry.Key * entry.Value;
}
}
return sum;
}
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));
}
}
|
Javascript
<script>
function findSum(arr,N)
{
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);
}
}
let maxFreq = 0;
for (let [key, value] of mp.entries())
{
if (value > maxFreq)
{
maxFreq = value;
}
}
let sum = 0;
for (let [key, value] of mp.entries())
{
if (value == maxFreq)
{
sum += key * value;
}
}
return sum;
}
let arr=[ 1, 1, 2, 2, 2, 2, 3, 3, 3, 3 ];
let N = arr.length;
document.write(findSum(arr, N));
</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
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