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<bits/stdc++.h>
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[0]);

        // Function calling
        cout << getOddOccurrence(arr, n);

        return 0;
    }


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 


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 :

//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<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

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 <bits/stdc++.h>
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[0]);
    
    // Function calling
    cout << getOddOccurrence(ar, n);
    
    return 0;
}

C

    
// C program to find the element
// occurring odd number of times
#include <stdio.h>

// 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[0]);
    
    // 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)

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