Find the Number Occurring Odd Number of Times

Given an array of positive integers. All numbers occur even number of times except one number which occurs 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 inner loop counts number of occurrences of the element picked by outer loop. Time complexity of this solution is O(n2).

Below is the implementation of the brute force approach :

C++

 // C++ program to find the element  // occurring odd number of times #include using namespace std;    // Funtion to find the element  // occurring odd number of times 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 int main()     {         int arr[] = { 2, 3, 5, 4, 5, 2,                       4, 3, 5, 2, 4, 4, 2 };         int n = sizeof(arr) / sizeof(arr);            // Function calling         cout << getOddOccurrence(arr, n);            return 0;     }

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Java

 // Java program to find the element occurring // odd number of times class OddOccurrence {            // funtion 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

Python3

 # Python program to find the element occurring # odd number of times        # funtion to find the element occurring odd # number of times def 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 code  arr = [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#

 // C# program to find the element  // occurring odd number of times using System;    class GFG {     // Funtion 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



Output :

5

A Better Solution is to use Hashing. Use array elements as key and their counts as value. Create an empty hash table. One by one traverse the given array elements and store counts. Time complexity of this solution is O(n). But it requires extra space for hashing.

Program :

C++

 // C++ program to find the element   // occurring odd number of times  #include using namespace std;    // funtion to find the element  // occurring odd number of times  int getOddOccurrence(int arr[],int size) {            // Defining HashMap in C++     unordered_map 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 code int main()  {      int arr[] = { 2, 3, 5, 4, 5, 2, 4,                     3, 5, 2, 4, 4, 2 };      int n = sizeof(arr) / sizeof(arr);             // Function calling      cout << getOddOccurrence(arr, n);         return 0;  }     // This code is contributed by codeMan_d.

Java

 // Java program to find the element occurring odd  // number of times import java.io.*; import java.util.HashMap;    class OddOccurrence  {     // funtion to find the element occurring odd     // number of times     static int getOddOccurrence(int arr[], int n)     {         HashMap 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

 # Python3 program to find the element   # occurring odd number of times      # funtion to find the element  # occurring odd number of times  def 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 code arr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2] n = len(arr)     # Function calling  print(getOddOccurrence(arr, n))     # This code is contributed by mohit kumar

Output :

5

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring element. Please note that XOR of two elements is 0 if both elements are same and XOR of a number x with 0 is x.

Below is the implementation of the above approach.

C++

 // C++ program to find the element // occurring odd number of times #include using namespace std;    // Function to find element occurring // odd number of times int getOddOccurrence(int ar[], int ar_size) {     int res = 0;      for (int i = 0; i < ar_size; i++)              res = res ^ ar[i];            return res; }    /* Diver 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);            // Function calling     cout << getOddOccurrence(ar, n);            return 0; }

C

 // C program to find the element // occurring odd number of times #include    // Function to find element occurring // odd number of times int getOddOccurrence(int ar[], int ar_size) {     int res = 0;      for (int i = 0; i < ar_size; i++)              res = res ^ ar[i];            return res; }    /* Diver 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);            // Function calling     printf("%d", getOddOccurrence(ar, n));     return 0; }

Java

 //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

 # 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 array arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2]    print "%d" % getOddOccurrence(arr)

C#

 // C# program to find the element // occurring odd number of times using System;    class GFG {     // Funtion 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



Output :

5

Time Complexity:
O(n)