Java Program to 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 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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Java

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

Output:

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

// Java program to find the element occurring odd 
// number of times 
import 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 

Output:

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 

Output:

Please refer complete article on Find the Number Occurring Odd Number of Times for more details!

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Java

Java

Java

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