# Find the element having different frequency than other array elements

Given an array of N integers. Each element in the array occurs the 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 the others.

Implementation:

## C++

 `// C++ program to find the element having``// different frequency than other array``// elements having same frequency``#include ``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;``}`

## Java

 `import` `java.util.*;` `public` `class` `DifferentFrequencyElement {``    ``// Function to find the element having different frequency``    ``// from other array elements with the same frequency``    ``static` `int` `findElement(``int``[] arr) {``        ``// Store frequencies of elements``        ``HashMap freq = ``new` `HashMap<>();``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``// Increase the value by 1 for every time the element occurs in the array``            ``freq.put(arr[i], freq.getOrDefault(arr[i], ``0``) + ``1``);``        ``}` `        ``// Initialize variables to store frequencies and elements``        ``int` `firstFreq = -``1``, secondFreq = -``1``, thirdFreq = -``1``;``        ``int` `firstElement = -``1``, secondElement = -``1``, thirdElement = -``1``;` `        ``for` `(Map.Entry entry : freq.entrySet()) {``            ``int` `element = entry.getKey();``            ``int` `frequency = entry.getValue();` `            ``// Update the variables based on frequency``            ``if` `(firstFreq == -``1``) {``                ``firstFreq = frequency;``                ``firstElement = element;``            ``} ``else` `if` `(frequency == firstFreq) {``                ``// If the current frequency is the same as the first frequency, update secondFreq and secondElement``                ``secondFreq = frequency;``                ``secondElement = element;``            ``} ``else` `if` `(frequency == secondFreq) {``                ``// If the current frequency is the same as the second frequency, update thirdFreq and thirdElement``                ``thirdFreq = frequency;``                ``thirdElement = element;``            ``} ``else` `{``                ``// If the frequency is different from both firstFreq and secondFreq, return the element``                ``return` `element;``            ``}``        ``}` `        ``// If the third frequency is different, return the third element``        ``if` `(secondFreq == firstFreq && secondFreq != thirdFreq)``            ``return` `thirdElement;``        ``if` `(secondFreq == thirdFreq && secondFreq != firstFreq)``            ``return` `firstElement;``        ``if` `(firstFreq == thirdFreq && secondFreq != firstFreq)``            ``return` `secondElement;` `        ``// We reach here when the first three frequencies are the same``        ``for` `(Map.Entry entry : freq.entrySet()) {``            ``if` `(entry.getValue() != firstFreq)``                ``return` `entry.getKey();``        ``}` `        ``return` `-``1``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = { ``0``, ``1``, ``2``, ``4``, ``4` `};``        ``int` `result = findElement(arr);``        ``System.out.println(result);``    ``}``}`

## Python3

 `# Python program to find the element having ``# different frequency than other array ``# elements having same frequency ` `# Function for above implementation``def` `findElement(arr, n) :``    ` `    ``# Empty dictionary to hold the values``    ``freq ``=` `{}``    ` `    ``# Initialization of frequencies of each ``    ``# element to 0``    ``for` `i ``in` `range``(``0``, n) :``        ` `        ``freq[arr[i]] ``=` `0``        ` `    ``# Count of frequencies of elements``    ``for` `i ``in` `range``(``0``, n) :``        ` `        ``freq[arr[i]] ``=` `freq[arr[i]] ``+` `1``     ` `    ``# Storing the first value of dictionary``    ``trd_ele ``=` `freq[``0``]``    ` `    ``# Variable to hold the final result``    ``position ``=` `-``1``    ` `    ``# Following loop iterates through the dictionary``    ``# and checks if frequencies are different ``    ``# from the frequency of the first element``    ``for` `i ``in` `freq :``        ` `        ``flag ``=` `freq[i]``        ` `        ``if` `trd_ele !``=` `flag :``            ` `            ``# Difference has been detected``            ``position ``=` `i``            ``break``            ` `    ``# Following lines of code checks if the first``    ``# element is itself the required anomaly by ``    ``# comparing the frequencies of first 3 elements``    ``fst_ele ``=` `freq[``1``]``    ``sec_ele ``=` `freq[``2``]``    ` `    ``if` `trd_ele !``=` `fst_ele :``        ` `        ``if` `trd_ele !``=` `sec_ele :``            ` `            ``for` `i ``in` `freq :``                ` `                ``# First element is the desired result``                ``position ``=` `i``                ``break``    ` `    ``# Final result is returned``    ``return` `position``    `  `# Driver code``arr ``=` `[ ``0``, ``1``, ``2``, ``4``, ``4` `]``# Variable to store length of array``n ``=` `len``(arr)``print` `(findElement(arr, n))` `# This code is contributed by Pratik Basu `

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `DifferentFrequencyElement``{``    ``// Function to find the element having different frequency``    ``// from other array elements with the same frequency``    ``static` `int` `FindElement(``int``[] arr)``    ``{``        ``// Store frequencies of elements``        ``Dictionary<``int``, ``int``> freq = ``new` `Dictionary<``int``, ``int``>();``        ``foreach` `(``int` `num ``in` `arr)``        ``{``            ``// Increase the value by 1 for every time the element occurs in the array``            ``if` `(freq.ContainsKey(num))``            ``{``                ``freq[num]++;``            ``}``            ``else``            ``{``                ``freq[num] = 1;``            ``}``        ``}` `        ``// Initialize variables to store frequencies and elements``        ``int` `firstFreq = -1, secondFreq = -1, thirdFreq = -1;``        ``int` `firstElement = -1, secondElement = -1, thirdElement = -1;` `        ``foreach` `(KeyValuePair<``int``, ``int``> entry ``in` `freq)``        ``{``            ``int` `element = entry.Key;``            ``int` `frequency = entry.Value;` `            ``// Update the variables based on frequency``            ``if` `(firstFreq == -1)``            ``{``                ``firstFreq = frequency;``                ``firstElement = element;``            ``}``            ``else` `if` `(frequency == firstFreq)``            ``{``                ``// If the current frequency is the same as the first frequency, update secondFreq and secondElement``                ``secondFreq = frequency;``                ``secondElement = element;``            ``}``            ``else` `if` `(frequency == secondFreq)``            ``{``                ``// If the current frequency is the same as the second frequency, update thirdFreq and thirdElement``                ``thirdFreq = frequency;``                ``thirdElement = element;``            ``}``            ``else``            ``{``                ``// If the frequency is different from both firstFreq and secondFreq, return the element``                ``return` `element;``            ``}``        ``}` `        ``// If the third frequency is different, return the third element``        ``if` `(secondFreq == firstFreq && secondFreq != thirdFreq)``            ``return` `thirdElement;``        ``if` `(secondFreq == thirdFreq && secondFreq != firstFreq)``            ``return` `firstElement;``        ``if` `(firstFreq == thirdFreq && secondFreq != firstFreq)``            ``return` `secondElement;` `        ``// We reach here when the first three frequencies are the same``        ``foreach` `(KeyValuePair<``int``, ``int``> entry ``in` `freq)``        ``{``            ``if` `(entry.Value != firstFreq)``                ``return` `entry.Key;``        ``}` `        ``return` `-1;``    ``}` `    ``static` `void` `Main()``    ``{``        ``int``[] arr = { 0, 1, 2, 4, 4 };``        ``int` `result = FindElement(arr);``        ``Console.WriteLine(result);``    ``}``}`

## Javascript

 `// JavaScript implementation``function` `findElement(arr, n) {` `    ``// Empty object to hold the values``    ``let freq = {};``    ` `    ``// Initialization of frequencies of each ``    ``// element to 0``    ``for``(let i = 0; i < n; i++) {``        ``freq[arr[i]] = 0;``    ``}``    ` `    ``// Count of frequencies of elements``    ``for``(let i = 0; i < n; i++) {``        ``freq[arr[i]] = freq[arr[i]] + 1;``    ``}``    ` `    ``// Storing the first value of object``    ``let trd_ele = freq[0];``    ` `    ``// Variable to hold the final result``    ``let position = -1;``    ` `    ``// Following loop iterates through the object``    ``// and checks if frequencies are different ``    ``// from the frequency of the first element``    ``for``(let i ``in` `freq) {``        ``let flag = freq[i];``        ``if` `(trd_ele !== flag) {``            ``// Difference has been detected``            ``position = i;``            ``break``;``        ``}``    ``}``    ` `    ``// Following lines of code checks if the first``    ``// element is itself the required anomaly by ``    ``// comparing the frequencies of first 3 elements``    ``let fst_ele = freq[1];``    ``let sec_ele = freq[2];``    ` `    ``if` `(trd_ele !== fst_ele) {``        ``if` `(trd_ele !== sec_ele) {``            ``for``(let i ``in` `freq) {``                ``// First element is the desired result``                ``position = i;``                ``break``;``            ``}``        ``}``    ``}``    ` `    ``// Final result is returned``    ``return` `position;` `}` `// Driver code``let arr = [ 0, 1, 2, 4, 4 ];``// Variable to store length of array``let n = arr.length;``console.log(findElement(arr, n));` `// This code is contributed by  phasing17`

Output
```4

```

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(n)

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