Find the Number Occurring Odd Number of Times - GeeksforGeeks

09 Dec, 2021
Given an array of positive integers. All numbers occur an even number of times except one number which occurs an odd number of times. Find the number in O(n) time & constant space.
Examples : 
Input : arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5
A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop. The time complexity of this solution is O(n2).
Below is the implementation of the brute force approach : 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find the element// occurring odd number of times#include<bits/stdc++.h>using namespace std; // Function to find the element// occurring odd number of timesint getOddOccurrence(int arr[], int arr_size){    for (int i = 0; i < arr_size; i++) {                 int count = 0;                 for (int j = 0; j < arr_size; j++)        {            if (arr[i] == arr[j])                count++;        }        if (count % 2 != 0)            return arr[i];    }    return -1;} // driver codeint main()    {        int arr[] = { 2, 3, 5, 4, 5, 2,                      4, 3, 5, 2, 4, 4, 2 };        int n = sizeof(arr) / sizeof(arr[0]);         // Function calling        cout << getOddOccurrence(arr, n);         return 0;    }
// Java program to find the element occurring// odd number of timesclass OddOccurrence {         // function to find the element occurring odd    // number of times    static int getOddOccurrence(int arr[], int arr_size)    {        int i;        for (i = 0; i < arr_size; i++) {            int count = 0;            for (int j = 0; j < arr_size; j++) {                if (arr[i] == arr[j])                    count++;            }            if (count % 2 != 0)                return arr[i];        }        return -1;    }         // driver code    public static void main(String[] args)    {        int arr[] = new int[]{ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };        int n = arr.length;        System.out.println(getOddOccurrence(arr, n));    }}// This code has been contributed by Kamal Rawal
# Python program to find the element occurring# odd number of times     # function to find the element occurring odd# number of timesdef getOddOccurrence(arr, arr_size):         for i in range(0,arr_size):        count = 0        for j in range(0, arr_size):            if arr[i] == arr[j]:                count+=1                     if (count % 2 != 0):            return arr[i]             return -1          # driver codearr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]n = len(arr)print(getOddOccurrence(arr, n)) # This code has been contributed by# Smitha Dinesh Semwal
// C# program to find the element// occurring odd number of timesusing System; class GFG{    // Function to find the element    // occurring odd number of times    static int getOddOccurrence(int []arr, int arr_size)    {        for (int i = 0; i < arr_size; i++) {            int count = 0;                         for (int j = 0; j < arr_size; j++) {                if (arr[i] == arr[j])                    count++;            }            if (count % 2 != 0)                return arr[i];        }        return -1;    }         // Driver code    public static void Main()    {        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };        int n = arr.Length;        Console.Write(getOddOccurrence(arr, n));    }} // This code is contributed by Sam007
<?php// PHP program to find the// element occurring odd// number of times // Function to find the element// occurring odd number of timesfunction getOddOccurrence(&$arr, $arr_size){    $count = 0;    for ($i = 0;         $i < $arr_size; $i++)    {                 for ($j = 0;             $j < $arr_size; $j++)        {            if ($arr[$i] == $arr[$j])                $count++;        }        if ($count % 2 != 0)            return $arr[$i];    }    return -1;} // Driver code$arr = array(2, 3, 5, 4, 5, 2,             4, 3, 5, 2, 4, 4, 2);$n = sizeof($arr); // Function callingecho(getOddOccurrence($arr, $n)); // This code is contributed// by Shivi_Aggarwal?>
<script> // Javascript program to find the element// occurring odd number of times // Function to find the element// occurring odd number of timesfunction getOddOccurrence(arr, arr_size){    for (let i = 0; i < arr_size; i++) {                 let count = 0;                 for (let j = 0; j < arr_size; j++)        {            if (arr[i] == arr[j])                count++;        }        if (count % 2 != 0)            return arr[i];    }    return -1;} // driver code         let arr = [ 2, 3, 5, 4, 5, 2,                    4, 3, 5, 2, 4, 4, 2 ];        let n = arr.length;         // Function calling        document.write(getOddOccurrence(arr, n)); // This code is contributed by Mayank Tyagi </script>
Output :  
5
Time Complexity: O(n^2)
Auxiliary Space: O(1)
A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts. The time complexity of this solution is O(n). But it requires extra space for hashing.
Program : 
C++
Java
Python3
C#
Javascript
// C++ program to find the element // occurring odd number of times#include <bits/stdc++.h>using namespace std; // function to find the element// occurring odd number of timesint getOddOccurrence(int arr[],int size){         // Defining HashMap in C++    unordered_map<int, int> hash;     // Putting all elements into the HashMap    for(int i = 0; i < size; i++)    {        hash[arr[i]]++;    }    // Iterate through HashMap to check an element    // occurring odd number of times and return it    for(auto i : hash)    {        if(i.second % 2 != 0)        {            return i.first;        }    }    return -1;} // Driver codeint main(){    int arr[] = { 2, 3, 5, 4, 5, 2, 4,                    3, 5, 2, 4, 4, 2 };    int n = sizeof(arr) / sizeof(arr[0]);         // Function calling    cout << getOddOccurrence(arr, n);     return 0;} // This code is contributed by codeMan_d.
// Java program to find the element occurring odd// number of timesimport java.io.*;import java.util.HashMap; class OddOccurrence{    // function to find the element occurring odd    // number of times    static int getOddOccurrence(int arr[], int n)    {        HashMap<Integer,Integer> hmap = new HashMap<>();                 // Putting all elements into the HashMap        for(int i = 0; i < n; i++)        {            if(hmap.containsKey(arr[i]))            {                int val = hmap.get(arr[i]);                                         // If array element is already present then                // increase the count of that element.                hmap.put(arr[i], val + 1);            }            else                                 // if array element is not present then put                // element into the HashMap and initialize                // the count to one.                hmap.put(arr[i], 1);        }         // Checking for odd occurrence of each element present        // in the HashMap        for(Integer a:hmap.keySet())        {            if(hmap.get(a) % 2 != 0)                return a;        }        return -1;    }             // driver code       public static void main(String[] args)    {        int arr[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};        int n = arr.length;        System.out.println(getOddOccurrence(arr, n));    }}// This code is contributed by Kamal Rawal
# Python3 program to find the element # occurring odd number of times  # function to find the element# occurring odd number of timesdef getOddOccurrence(arr,size):          # Defining HashMap in C++    Hash=dict()      # Putting all elements into the HashMap    for i in range(size):        Hash[arr[i]]=Hash.get(arr[i],0) + 1;         # Iterate through HashMap to check an element    # occurring odd number of times and return it    for i in Hash:         if(Hash[i]% 2 != 0):            return i    return -1   # Driver codearr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2]n = len(arr)  # Function callingprint(getOddOccurrence(arr, n)) # This code is contributed by mohit kumar
// C# program to find the element occurring odd// number of timesusing System;using System.Collections.Generic; public class OddOccurrence{    // function to find the element occurring odd    // number of times    static int getOddOccurrence(int []arr, int n)    {        Dictionary<int,int> hmap = new Dictionary<int,int>();                 // Putting all elements into the HashMap        for(int i = 0; i < n; i++)        {            if(hmap.ContainsKey(arr[i]))            {                int val = hmap[arr[i]];                                         // If array element is already present then                // increase the count of that element.                hmap.Remove(arr[i]);                hmap.Add(arr[i], val + 1);            }            else                                 // if array element is not present then put                // element into the HashMap and initialize                // the count to one.                hmap.Add(arr[i], 1);        }         // Checking for odd occurrence of each element present        // in the HashMap        foreach(KeyValuePair<int, int> entry in hmap)        {            if(entry.Value % 2 != 0)            {                return entry.Key;            }        }        return -1;    }             // Driver code    public static void Main(String[] args)    {        int []arr = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};        int n = arr.Length;        Console.WriteLine(getOddOccurrence(arr, n));    }} // This code is contributed by Princi Singh
<script>// Javascript program to find the element occurring odd// number of times         // function to find the element occurring odd    // number of times    function getOddOccurrence(arr,n)    {        let hmap = new Map();                  // Putting all elements into the HashMap        for(let i = 0; i < n; i++)        {            if(hmap.has(arr[i]))            {                let val = hmap.get(arr[i]);                // If array element is already present then                // increase the count of that element.                hmap.set(arr[i], val + 1);            }            else            {                    // if array element is not present then put                // element into the HashMap and initialize                // the count to one.                hmap.set(arr[i], 1);            }        }                   // Checking for odd occurrence of each element present        // in the HashMap        for(let [key, value] of hmap.entries())        {            //document.write(hmap[a]+"<br>")            if(hmap.get(key) % 2 != 0)                return key;        }        return -1;    }         // driver code      let arr=[2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2];    let n = arr.length;    document.write(getOddOccurrence(arr, n));              // This code is contributed by unknown2108</script>
Output : 
5
Time Complexity: O(n)
Auxiliary Space: O(n)
The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements. Please note that the XOR of two elements is 0 if both elements are the same and the XOR of a number x with 0 is x.
Below is the implementation of the above approach.  
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program to find the element// occurring odd number of times#include <bits/stdc++.h>using namespace std; // Function to find element occurring// odd number of timesint getOddOccurrence(int ar[], int ar_size){    int res = 0;    for (int i = 0; i < ar_size; i++)            res = res ^ ar[i];         return res;} /* Driver function to test above function */int main(){    int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};    int n = sizeof(ar)/sizeof(ar[0]);         // Function calling    cout << getOddOccurrence(ar, n);         return 0;}
// C program to find the element// occurring odd number of times#include <stdio.h> // Function to find element occurring// odd number of timesint getOddOccurrence(int ar[], int ar_size){    int res = 0;    for (int i = 0; i < ar_size; i++)            res = res ^ ar[i];         return res;} /* Driver function to test above function */int main(){    int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};    int n = sizeof(ar) / sizeof(ar[0]);         // Function calling    printf("%d", getOddOccurrence(ar, n));    return 0;}
//Java program to find the element occurring odd number of times class OddOccurance{    int getOddOccurrence(int ar[], int ar_size)    {        int i;        int res = 0;        for (i = 0; i < ar_size; i++)        {            res = res ^ ar[i];        }        return res;    }     public static void main(String[] args)    {        OddOccurance occur = new OddOccurance();        int ar[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};        int n = ar.length;        System.out.println(occur.getOddOccurrence(ar, n));    }}// This code has been contributed by Mayank Jaiswal
# Python program to find the element occurring odd number of times def getOddOccurrence(arr):     # Initialize result    res = 0         # Traverse the array    for element in arr:        # XOR with the result        res = res ^ element     return res # Test arrayarr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2] print("%d" % getOddOccurrence(arr))
// C# program to find the element// occurring odd number of timesusing System; class GFG{    // Function to find the element    // occurring odd number of times    static int getOddOccurrence(int []arr, int arr_size)    {        int res = 0;        for (int i = 0; i < arr_size; i++)        {            res = res ^ arr[i];        }        return res;    }         // Driver code    public static void Main()    {        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };        int n = arr.Length;        Console.Write(getOddOccurrence(arr, n));    }} // This code is contributed by Sam007
<?php// PHP program to find the// element occurring odd// number of times // Function to find element// occurring odd number of timesfunction getOddOccurrence(&$ar, $ar_size){    $res = 0;    for ($i = 0; $i < $ar_size; $i++)            $res = $res ^ $ar[$i];         return $res;} // Driver Code$ar = array(2, 3, 5, 4, 5, 2,            4, 3, 5, 2, 4, 4, 2);$n = sizeof($ar); // Function callingecho(getOddOccurrence($ar, $n));     // This code is contributed// by Shivi_Aggarwal?>
<script> // JavaScript program to find the element// occurring odd number of times // Function to find the element// occurring odd number of timesfunction getOddOccurrence( ar, ar_size){    let res = 0;         for (let i = 0; i < ar_size; i++)            res = res ^ ar[i];         return res;}     // driver code     let arr = [ 2, 3, 5, 4, 5, 2, 4,                3, 5, 2, 4, 4, 2 ];    let n = arr.length;     // Function calling    document.write(getOddOccurrence(arr, n));  </script>
Output :
5
Time Complexity: O(n)
Auxiliary Space: O(1)
Count the frequencies of every element using the Counter function
Traverse in frequency dictionary
Check which element has an odd frequency.
Print that element and break the loop
Below is the implementation:
Python3
# importing counter from collectionsfrom collections import Counter # Python3 implementation to find# odd frequency elementdef oddElement(arr, n):     # Calculating frequencies using Counter    count_map = Counter(arr)     for i in range(0, n):         # If count of element is odd we return        if (count_map[arr[i]] % 2 != 0):            return arr[i]  # Driver Codeif __name__ == "__main__":     arr = [1, 1, 3, 3, 5, 6, 6]    n = len(arr)    print(oddElement(arr, n)) # This code is contributed by vikkycirus
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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