# 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)

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