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 cde 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` |

**Output:**

30

**Time Complexity**: O(N), where N is the number of elements in the array.

**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:**

## 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 freeq and print all` ` ` `# sum all elements with odd frequency` ` ` `# multiplieed 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` |

**Output:**

60

**Time Complexity:** O(N), where N is the number of elements in the array.

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