# Find the element having different frequency than other array elements

Given an array of N integers. Each element in the array occurs same number of times except one element. The task is to find this element.

**Examples:**

Input :arr[] = {1, 1, 2, 2, 3}Output :3Input :arr[] = {0, 1, 2, 4, 4}Output :4

The idea is to use a hash table **freq** to store the frequencies of given elements. Once we have frequencies in the hash table, we can traverse the table to find the only value which is different from others.

Below is the implementation of the above idea :

`// C++ program to find the element having ` `// different frequency than other array ` `// elements having same frequency ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the element having ` `// different frequency from other array ` `// elements with same frequency ` `int` `findElement(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// Store frequencies of elements ` ` ` `unordered_map<` `int` `, ` `int` `> freq; ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// increase the value by 1 for every ` ` ` `// time the element occurs in an array ` ` ` `freq[arr[i]]++; ` ` ` `} ` ` ` ` ` `// Below code is used find the only different ` ` ` `// value in freq. ` ` ` `auto` `it = freq.begin(); ` ` ` `int` `fst_fre = it->second, fst_ele = it->first; ` ` ` `if` `(freq.size() <= 2) ` ` ` `return` `fst_fre; ` ` ` `it++; ` ` ` `int` `sec_fre = it->second, sec_ele = it->first; ` ` ` `it++; ` ` ` `int` `trd_fre = it->second, trd_ele = it->first; ` ` ` `if` `(sec_fre == fst_fre && sec_fre != trd_fre) ` ` ` `return` `trd_ele; ` ` ` `if` `(sec_fre == trd_fre && sec_fre != fst_fre) ` ` ` `return` `fst_ele; ` ` ` `if` `(fst_fre == trd_fre && sec_fre != fst_fre) ` ` ` `return` `sec_ele; ` ` ` ` ` `// We reach here when first three frequencies are same ` ` ` `it++; ` ` ` `for` `(; it != freq.end(); it++) { ` ` ` `if` `(it->second != fst_fre) ` ` ` `return` `it->first; ` ` ` `} ` ` ` ` ` `return` `-1; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 0, 1, 2, 4, 4 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `cout << findElement(arr, n) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

4

Time Complexity : O(n)

Auxiliary Space : O(n)

## Recommended Posts:

- Find element in a sorted array whose frequency is greater than or equal to n/2.
- Find frequency of each element in a limited range array in less than O(n) time
- Maximum difference between frequency of two elements such that element having greater frequency is also greater
- Replace each element by the difference of the total size of the array and frequency of that element
- Find Kth element in an array containing odd elements first and then even elements
- Find element in array that divides all array elements
- Find an array element such that all elements are divisible by it
- Find an element in an array such that elements form a strictly decreasing and increasing sequence
- Sum of all odd frequency elements in an array
- Cumulative frequency of count of each element in an unsorted array
- Frequency of each element of an array of small ranged values
- XOR of elements in an array having prime frequency
- Sorting Array Elements By Frequency | Set 3 (Using STL)
- Sum of elements in an array having prime frequency
- Array range queries for elements with frequency same as value

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.