Java Program to Find the Number Occurring Odd Number of Times
Last Updated :
09 Nov, 2023
Write a Python program for a given 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
Python Program to Find the Number Occurring Odd Number of Times using Nested Loop:
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.
Below is the implementation of the brute force approach :
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
Time Complexity: O(n^2)
Auxiliary Space: O(1)
Python Program to Find the Number Occurring Odd Number of Times using Hashing:
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.
Below is the implementation of the above approach:
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
Time Complexity: O(n)
Auxiliary Space: O(n)
Python Program to Find the Number Occurring Odd Number of Times using Bit Manipulation:
The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements.
Here ^ is the XOR operators;
Note :
x^0 = x
x^y=y^x (Commutative property holds)
(x^y)^z = x^(y^z) (Distributive property holds)
x^x=0
Below is the implementation of the above approach.
Java
//Java program to find the element occurring odd number of times
class OddOccurrence
{
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)
{
OddOccurrence occur = new OddOccurrence();
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
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Find the Number Occurring Odd Number of Times for more details!
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