# Sum of all odd frequency elements in an array

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; ` `} ` |

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## 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 */` |

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## 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. ` |

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**Output:**

30

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

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- Maximum product from array such that frequency sum of all repeating elements in product is less than or equal to 2 * k
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- Sum of all odd frequency elements in a Matrix
- Sum of all even frequency elements in Matrix
- Sort elements by frequency | Set 1
- Sort elements by frequency | Set 2
- Sum of all minimum frequency elements in Matrix
- Sum of all maximum frequency elements in Matrix

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