# Two odd occurring elements in an array where all other occur even times

• Difficulty Level : Easy
• Last Updated : 24 Nov, 2022

Given an array where all elements appear even number of times except two, print the two odd occurring elements. It may be assumed that the size of array is at-least two.

Examples:

```Input : arr[] = {2, 3, 8, 4, 4, 3, 7, 8}
Output : 2 7

Input : arr[] = {15, 10, 10, 50 7, 5, 5, 50, 50, 50, 50, 50}
Output : 7 15```

Simple solution :

Approach :

A simple solution is to use two nested loops. The outer loop traverses through all elements. The inner loop counts occurrences of the current element. We print the elements whose counts of occurrences are odd.

Below is the code of the given approach :

## C++

 `// CPP code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``#include ``using` `namespace` `std;` `void` `printOdds(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `count = 0;``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `(arr[i] == arr[j]) {``                ``count += 1;``            ``}``        ``}``        ``if` `(count % 2 != 0) {``            ``cout << arr[i]``                 ``<< ``" "``; ``// Print the elements that occur``                         ``// odd number of times in an array``        ``}``    ``}``    ``cout << endl;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 3, 4, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``// function call``    ``printOdds(arr, n);``    ``return` `0;``}` `// This code is contributed by Suruchi Kumari`

## C

 `// C code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``#include ` `void` `printOdds(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `count = 0;``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `(arr[i] == arr[j]) {``                ``count += 1;``            ``}``        ``}``        ``if` `(count % 2 != 0) {``            ``printf``(``                ``"%d "``,``                ``arr[i]); ``// Print the elements that occur``                         ``// odd number of times in an array``        ``}``    ``}``    ``printf``(``"\n"``);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 3, 4, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``// function call``    ``printOdds(arr, n);``    ``return` `0;``}` `// This code is contributed by Suruchi Kumari`

## Java

 `// Java program to find the maximum stolen value``public` `class` `GFG {` `  ``// calculate the maximum stolen value``  ``static` `void` `printOdds(``int``[] arr, ``int` `n)``  ``{``    ``for` `(``int` `i = ``0``; i < n; i++) {``      ``int` `count = ``0``;``      ``for` `(``int` `j = ``0``; j < n; j++) {``        ``if` `(arr[i] == arr[j]) {``          ``count += ``1``;``        ``}``      ``}``      ``if` `(count % ``2` `!= ``0``) {``        ``System.out.print(arr[i]+``" "``);``// Print the elements that occur``        ``// odd number of times in an array``      ``}``    ``}``    ``System.out.print(``"\n"``);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main (String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``3``, ``3``, ``4``, ``4``, ``5` `};``    ``int` `n = arr.length;` `    ``// function call``    ``printOdds(arr, n);``  ``}``}``// This code is contributed by adityapatil12`

## Python3

 `# python3 code for the above approach` `# Function to find and Replace in String``def` `printOdds(arr, n) :``    ` `    ``for` `i ``in` `range``(``0``,n) :``        ``count ``=` `0``        ``for` `j ``in` `range``(``0``,n) :``            ``if` `arr[i] ``=``=` `arr[j] :``                ``count ``+``=` `1``        ``if` `count ``%` `2` `!``=` `0` `:``            ``print``(arr[i],end``=``' '``) ``# Print the elements that occur``                                   ``# odd number of times in an array``  ` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``2``, ``3``, ``3``, ``4``, ``4``, ``5` `]``    ``n ``=` `len``(arr)``    ` `    ``#Function call``    ``printOdds(arr,n)` `# This code is contributed by adityapatil12`

## C#

 `// Include namespace system``using` `System;` `// C# program to find the maximum stolen value``public` `class` `GFG``{``    ``// calculate the maximum stolen value``    ``public` `static` `void` `printOdds(``int``[] arr, ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``var` `count = 0;``            ``for` `(``int` `j = 0; j < n; j++)``            ``{``                ``if` `(arr[i] == arr[j])``                ``{``                    ``count += 1;``                ``}``            ``}``            ``if` `(count % 2 != 0)``            ``{``                ``Console.Write(arr[i].ToString() + ``" "``);``            ``}``        ``}``        ``Console.Write(``"\n"``);``    ``}``    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = {2, 3, 3, 4, 4, 5};``        ``var` `n = arr.Length;``        ``// function call``        ``GFG.printOdds(arr, n);``    ``}``}` `// This code is contributed by aadityaburujwale.`

## Javascript

 `// JavaScript code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``function` `printOdds(arr, n)``{``    ``for` `(``var` `i = 0; i < n; i++) {``        ``let count = 0;``        ``for` `(``var` `j = 0; j < n; j++) {``            ``if` `(arr[i] == arr[j]) {``                ``count += 1;``            ``}``        ``}``        ``if` `(count % 2 != 0) {``            ``process.stdout.write(arr[i] + ``" "``);``                        ``// Print the elements that occur``                         ``// odd number of times in an array``        ``}``    ``}``    ``process.stdout.write(``"\n"``);``}` `// Driver code``let arr = [ 2, 3, 3, 4, 4, 5 ];``let n = arr.length;` `// function call``printOdds(arr, n);`  `// This code is contributed by phasing17`

Output

`2 5 `

Complexity Analysis:

• Time complexity : O(n^2)
• Auxiliary space : O(n)

A better solution is to use hashing. Time complexity of this solution is O(n) but it requires extra space.

We can construct a frequency hashmap, then iterate over all of its key – value pairs, and print all keys whose values (frequency) are odd.

Implementation:

## C++

 `// CPP code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``#include ``using` `namespace` `std;`  `void` `printOdds(``int` `arr[], ``int` `n)``{``    ``//declaring an unordered map``    ``unordered_map<``int``, ``int``> freq;``    ``//building the frequency table``    ``for` `(``int` `i = 0; i < n; i++)``        ``freq[arr[i]]++;``    ` `    ``//iterating over the map``    ``for` `(``auto``& it: freq) {``        ``//if the frequency is odd``        ``//print the element``        ``if` `(it.second % 2)``            ``cout << it.first << ``" "``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 3, 4, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``//function call``    ``printOdds(arr, n);``    ``return` `0;``}` `//this code is contributed by phasing17`

## Java

 `/*package whatever //do not write package name here */``import` `java.util.*;` `class` `GFG {``    ``static` `void` `printOdds(``int` `arr[], ``int` `n)``    ``{``        ``// declaring an unordered map``        ``TreeMap freq = ``new` `TreeMap<>(Collections.reverseOrder());``      ` `        ``// building the frequency table``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``freq.put(arr[i],``                     ``freq.getOrDefault(arr[i], ``0``) + ``1``);``        ``}` `        ``// iterating over the map``        ``for` `(``int` `i : freq.keySet()) {``            ``// if the frequency is odd``            ``// print the element``            ``if` `(freq.get(i) % ``2` `== ``1``)``                ``System.out.print(i + ``" "``);``        ``}``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``3``, ``3``, ``4``, ``4``, ``5` `};``        ``int` `n = arr.length;``      ` `        ``// function call``        ``printOdds(arr, n);``    ``}``}` `// This code is contributed by aadityaburujwale.`

## C#

 `// C# code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG {``  ``static` `void` `printOdds(``int``[] arr, ``int` `n)``  ``{``    ` `    ``// declaring an unordered map``    ``Dictionary<``int``, ``int``> freq``      ``= ``new` `Dictionary<``int``, ``int``>();``    ` `    ``// building the frequency table``    ``for` `(``int` `i = 0; i < n; i++) {``      ``if` `(freq.ContainsKey(arr[i])) {``        ``int` `val = freq[arr[i]];``        ``freq.Remove(arr[i]);``        ``freq.Add(arr[i], val + 1);``      ``}``      ``else` `{``        ``freq.Add(arr[i], 1);``      ``}``    ``}` `    ``// iterating over the map``    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `freq)``    ``{``      ` `      ``// if the frequency is odd``      ``// print the element``      ``if` `(entry.Value % 2 != 0)``        ``Console.Write(entry.Key + ``" "``);``    ``}``  ``}` `  ``// Driver code``  ``static` `public` `void` `Main()``  ``{``    ``int``[] arr = { 2, 3, 3, 4, 4, 5 };``    ``int` `n = 6;``    ` `    ``// function call``    ``printOdds(arr, n);``  ``}``}` `// This code is contributed by garg28harsh.`

## Javascript

 `// function code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``function` `printOdds(arr, n)``{``    ``// declaring an unordered map``    ``var` `mp = ``new` `Map();``    ` `    ``// building the frequency table``    ``for` `(``var` `i = 0; i < n; i++)``    ``{``        ``if``(mp.has(arr[i]))``            ``mp.set(arr[i], mp.get(arr[i])+1);``        ``else``            ``mp.set(arr[i], 1);``    ``}``    ``let el=[];``    ``mp.forEach(( key,value) => {``        ``if``(key%2!=0)``        ``{``            ``el.push(value);``        ``}``    ``});``     ``console.log(el);``}` `// Driver code` `    ``let arr = [2, 3, 3, 4, 4, 5 ];``    ``let n = arr.length;``    ` `    ``// function call``    ``printOdds(arr, n);` `// This code is contributed by garg28harsh.`

Output

`5 2 `

Complexity Analysis:

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

An efficient solution is to use bitwise operators. The idea is based on approach used in two missing elements and two repeating elements.

Implementation:

## C++

 `// CPP code to find two odd occurring elements``// in an array where all other elements appear``// even number of times.``#include ``using` `namespace` `std;` `void` `printOdds(``int` `arr[], ``int` `n)``{``    ``// Find XOR of all numbers``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``res = res ^ arr[i];` `    ``// Find a set bit in the XOR (We find``    ``// rightmost set bit here)``    ``int` `set_bit = res & (~(res - 1));` `    ``// Traverse through all numbers and``    ``// divide them in two groups``    ``// (i) Having set bit set at same``    ``//     position as the only set bit``    ``//     in set_bit``    ``// (ii) Having 0 bit at same position``    ``//      as the only set bit in set_bit``    ``int` `x = 0, y = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] & set_bit)``            ``x = x ^ arr[i];``        ``else``            ``y = y ^ arr[i];``    ``}` `    ``// XOR of two different sets are our``    ``// required numbers.``    ``cout << x << ``" "` `<< y;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 3, 4, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``printOdds(arr, n);``    ``return` `0;``}`

## Java

 `// Java code to find two``// odd occurring elements``// in an array where all``// other elements appear``// even number of times.` `class` `GFG``{``static` `void` `printOdds(``int` `arr[],``                      ``int` `n)``{``    ``// Find XOR of``    ``// all numbers``    ``int` `res = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``res = res ^ arr[i];` `    ``// Find a set bit in the``    ``// XOR (We find rightmost``    ``// set bit here)``    ``int` `set_bit = res &``                  ``(~(res - ``1``));` `    ``// Traverse through all``    ``// numbers and divide them``    ``// in two groups (i) Having``    ``// set bit set at same position``    ``// as the only set bit in``    ``// set_bit (ii) Having 0 bit at``    ``// same position as the only``    ``// set bit in set_bit``    ``int` `x = ``0``, y = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``if` `((arr[i] & set_bit) != ``0``)``            ``x = x ^ arr[i];``        ``else``            ``y = y ^ arr[i];``    ``}` `    ``// XOR of two different``    ``// sets are our required``    ``// numbers.``    ``System.out.println( x + ``" "` `+ y);``}` `// Driver code``public` `static` `void` `main(String [] args)``{``    ``int` `arr[] = { ``2``, ``3``, ``3``,``                  ``4``, ``4``, ``5` `};``    ``int` `n = arr.length;``    ``printOdds(arr, n);``}``}` `// This code is contributed by``// Smitha Dinesh Semwal`

## Python3

 `# Python 3 code to find two``# odd occurring elements in``# an array where all other``# elements appear even number``# of times.``def` `printOdds(arr, n):` `    ``# Find XOR of all numbers``    ``res ``=` `0``    ``for` `i ``in` `range``(``0``, n):``        ``res ``=` `res ^ arr[i]` `    ``# Find a set bit in``    ``# the XOR (We find``    ``# rightmost set bit here)``    ``set_bit ``=` `res & (~(res ``-` `1``))` `    ``# Traverse through all numbers``    ``# and divide them in two groups``    ``# (i) Having set bit set at``    ``# same position as the only set``    ``# bit in set_bit``    ``# (ii) Having 0 bit at same``    ``# position as the only set``    ``# bit in set_bit``    ``x ``=` `0``    ``y ``=` `0``    ``for` `i ``in` `range``(``0``, n):``        ``if` `(arr[i] & set_bit):``            ``x ``=` `x ^ arr[i]``        ``else``:``            ``y ``=` `y ^ arr[i]``    ` `    ``# XOR of two different``    ``# sets are our``    ``# required numbers.``    ``print``(x , y, end ``=` `"")` `# Driver code``arr ``=` `[``2``, ``3``, ``3``, ``4``, ``4``, ``5` `]``n ``=` `len``(arr)``printOdds(arr, n)` `# This code is contributed``# by Smitha`

## C#

 `// C# code to find two``// odd occurring elements``// in an array where all``// other elements appear``// even number of times.``using` `System;` `class` `GFG``{``static` `void` `printOdds(``int` `[]arr,``                      ``int` `n)``{``    ``// Find XOR of``    ``// all numbers``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``res = res ^ arr[i];` `    ``// Find a set bit in the``    ``// XOR (We find rightmost``    ``// set bit here)``    ``int` `set_bit = res &``               ``(~(res - 1));` `    ``// Traverse through all``    ``// numbers and divide them``    ``// in two groups (i) Having``    ``// set bit set at same position``    ``// as the only set bit in``    ``// set_bit (ii) Having 0 bit at``    ``// same position as the only``    ``// set bit in set_bit``    ``int` `x = 0, y = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `((arr[i] & set_bit) != 0)``            ``x = x ^ arr[i];``        ``else``            ``y = y ^ arr[i];``    ``}` `    ``// XOR of two different``    ``// sets are our required``    ``// numbers.``    ``Console.WriteLine(x + ``" "` `+ y);``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 2, 3, 3,``                  ``4, 4, 5 };``    ``int` `n = arr.Length;``    ``printOdds(arr, n);``}``}` `// This code is contributed by``// Akanksha Rai(Abby_akku)`

## PHP

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## Javascript

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Output

`5 2`

Complexity Analysis:

• Time Complexity : O(n)
• Auxiliary Space : O(1)

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