Given an array of integers containing duplicate elements. The task is to find the sum of all odd occurring elements in the given array. That is the sum of all such elements whose frequency is odd in the array.
Examples:
Input : arr[] = {1, 1, 2, 2, 3, 3, 3} Output : 9 The odd occurring element is 3, and it's number of occurrence is 3. Therefore sum of all 3's in the array = 9. Input : arr[] = {10, 20, 30, 40, 40} Output : 60 Elements with odd frequency are 10, 20 and 30. Sum = 60.
Approach:
- Traverse the array and use a 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 elements and check if it is odd, if it is odd, then add this element to sum.
Below is the implementation of the above approach:
C++
// CPP program to find the sum of all odd // occurring elements in an array #include <bits/stdc++.h> using namespace std;
// Function to find the sum of all odd // 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]]++;
// variable to store sum of all
// odd occurring elements
int sum = 0;
// loop to iterate through map
for ( auto itr = mp.begin(); itr != mp.end(); itr++) {
// check if frequency is odd
if (itr->second % 2 != 0)
sum += (itr->first) * (itr->second);
}
return sum;
} // Driver Code int main()
{ int arr[] = { 10, 20, 20, 10, 40, 40, 10 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << findSum(arr, N);
return 0;
} |
Java
// Java program to find the sum of all odd // occurring elements in an array import java.util.*;
class GFG
{ // Function to find the sum of all odd // 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 );
// variable to store sum of all
// odd occurring elements
int sum = 0 ;
// loop to iterate through map
for (Map.Entry<Integer,Integer> entry : mp.entrySet())
{
// check if frequency is odd
if (entry.getValue() % 2 != 0 )
sum += (entry.getKey()) * (entry.getValue());
}
return sum;
} // Driver Code public static void main(String args[])
{ int arr[] = { 10 , 20 , 20 , 10 , 40 , 40 , 10 };
int N = arr.length;
System.out.println(findSum(arr, N));
} } /* This code is contributed by PrinciRaj1992 */ |
Python3
# Function to find sum of all odd # occurring elements in an array import collections
def findsum(arr, N):
# Store frequencies of elements
# of an array in dictionary
mp = collections.defaultdict( int )
for i in range (N):
mp[arr[i]] + = 1
# Variable to store sum of all
# odd occurring elements
sum = 0
# loop to iterate through dictionary
for i in mp:
# Check if frequency is odd
if (mp[i] % 2 ! = 0 ):
sum + = (i * mp[i])
return sum
# Driver Code arr = [ 10 , 20 , 20 , 10 , 40 , 40 , 10 ]
N = len (arr)
print (findsum(arr, N))
# This code is contributed # by HardeepSingh. |
C#
// C# program to find the sum of all odd // occurring elements in an array using System;
using System.Collections.Generic;
class GFG
{ // Function to find the sum of all odd
// occurring elements in an array
public 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);
}
// variable to store sum of all
// odd occurring elements
int sum = 0;
// loop to iterate through map
foreach (KeyValuePair< int ,
int > entry in mp)
{
// check if frequency is odd
if (entry.Value % 2 != 0)
sum += entry.Key * entry.Value;
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int [] arr = { 10, 20, 20, 10, 40, 40, 10 };
int n = arr.Length;
Console.WriteLine(findSum(arr, n));
}
} // This code is contributed by // sanjeev2552 |
Javascript
<script> // Javascript program to find the sum of all odd // occurring elements in an array // Function to find the sum of all odd // 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)
}
}
// variable to store sum of all
// odd occurring elements
let sum = 0;
// loop to iterate through map
for (let itr of mp) {
// check if frequency is odd
if (itr[1] % 2 != 0)
sum += (itr[0]) * (itr[1]);
}
return sum;
} // Driver Code let arr = [10, 20, 20, 10, 40, 40, 10]; let N = arr.length document.write(findSum(arr, N)); // This code is contributed by gfgking. </script> |
Output
30
complexity Analysis:
- Time Complexity: O(N), where N is the number of elements in the array.
- Auxiliary Space: O(N)
Method 2:Using Built in python functions:
- Count the frequencies of every element using Counter function
- Traverse the frequency dictionary and sum all the elements with occurrence odd frequency multiplied by its frequency.
Below is the implementation:
C++
#include <iostream> #include <unordered_map> using namespace std;
void sumOdd( int arr[], int n)
{ // Counting frequency of every element using unordered_map
unordered_map< int , int > freq;
for ( int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// Initializing sum to 0
int sum = 0;
// Traverse the frequency and sum all elements
// with odd frequency multiplied by its frequency
for ( auto it : freq) {
if (it.second % 2 != 0) {
sum += it.first * it.second;
}
}
cout << sum << endl;
} // Driver code int main() {
int arr[] = {10, 20, 30, 40, 40};
int n = sizeof (arr) / sizeof (arr[0]);
sumOdd(arr, n);
} |
Python3
# Python3 implementation of the above approach from collections import Counter
def sumOdd(arr, n):
# Counting frequency of every
# element using Counter
freq = Counter(arr)
# Initializing sum 0
sum = 0
# Traverse the frequency and print all
# sum all elements with odd frequency
# multiplied by its frequency
for it in freq:
if freq[it] % 2 ! = 0 :
sum = sum + it * freq[it]
print ( sum )
# Driver code arr = [ 10 , 20 , 30 , 40 , 40 ]
n = len (arr)
sumOdd(arr, n) # This code is contributed by vikkycirus |
C#
// C# implementation of the above approach using System;
using System.Collections.Generic;
using System.Linq;
class MainClass {
// Function to return the sum of elements
// in an array having odd frequency
public static void sumOdd( int [] arr, int n)
{
// Dictionary is used to calculate frequency of
// elements of array
Dictionary< int , int > freq
= arr.GroupBy(x => x).ToDictionary(
x => x.Key, x => x.Count());
int sum = 0;
// Traverse the dictionary
foreach (KeyValuePair< int , int > entry in freq)
{
// Calculate the sum of elements
// having odd frequency
// multiplied by its frequency
if (entry.Value % 2 != 0) {
sum += entry.Key * entry.Value;
}
}
Console.WriteLine(sum);
}
// Driver code
public static void Main()
{
int [] arr = { 10, 20, 30, 40, 40 };
int n = arr.Length;
sumOdd(arr, n);
}
} // This code is contributed by phasing17 |
Javascript
function sumOdd(arr, n) {
// Counting frequency of every
// element using Map
let freq = new Map();
for (let i=0; i<n; i++){
if (freq.has(arr[i])){
freq.set(arr[i], freq.get(arr[i])+1);
} else {
freq.set(arr[i], 1);
}
}
// Initializing sum to 0
let sum = 0;
// Traverse the frequency and sum all elements
// with odd frequency multiplied by its frequency
for (let [key, value] of freq){
if (value % 2 !== 0){
sum += key*value;
}
}
console.log(sum);
} // Driver code let arr = [10, 20, 30, 40, 40]; let n = arr.length; sumOdd(arr, n); |
Java
/*package whatever //do not write package name here */ import java.util.HashMap;
public class GFG {
static void sumOdd( int [] arr, int n) {
// Counting frequency of every element using HashMap
HashMap<Integer, Integer> freq = new HashMap<Integer, Integer>();
for ( int i = 0 ; i < n; i++) {
if (freq.containsKey(arr[i])) {
freq.put(arr[i], freq.get(arr[i]) + 1 );
} else {
freq.put(arr[i], 1 );
}
}
// Initializing sum to 0
int sum = 0 ;
// Traverse the frequency and sum all elements
// with odd frequency multiplied by its frequency
for (HashMap.Entry<Integer, Integer> entry : freq.entrySet()) {
if (entry.getValue() % 2 != 0 ) {
sum += entry.getKey() * entry.getValue();
}
}
System.out.println(sum);
}
// Driver code
public static void main(String[] args) {
int [] arr = { 10 , 20 , 30 , 40 , 40 };
int n = arr.length;
sumOdd(arr, n);
}
} |
Output
60
Complexity Analysis:
- Time Complexity: O(N), where N is the number of elements in the array.
- Auxiliary Space: O(N)