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.
// 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;
}
/* 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 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;
}
/* 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 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
# 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# program to find the element
// occurring odd number of times
using 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
// JavaScript program to find the element
// occurring odd number of times
// Function to find the element
// occurring odd number of times
function 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
console.log(getOddOccurrence(arr, n));
<?php
// PHP program to find the
// element occurring odd
// number of times
// Function to find element
// occurring odd number of times
function 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 calling
echo(getOddOccurrence($ar, $n));
// This code is contributed
// by Shivi_Aggarwal
?>
Output
5
Time Complexity: O(n)
Auxiliary Space: O(1)
Find the Number Occurring Odd Number of Times
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 : 3Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5
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