find the number occuring odd number of times

Hi,guys here is the simple solution(brute force approach). Time complexity:o(n^2) Space Complexity:o(1)

#include<bits/stdc++.h>
using namespace std;
int main(){
 int i,j,n;
 cout << "Enter the size of array:\n";
 cin >> n;

 int arr[n];
 //note that there is only one element which occure odd number of times
 cout << "Enter the elements of array:\n";
 for(i=0;i<n;i++){
  cin >> arr[i];
 }

 //brute force approach
 for(i=0;i<n;i++){
  int count=0;
  for(j=0;j<n;j++){
   if(arr[i]==arr[j]){
    count++;
   }
  }
  if(count%2==1){
   cout << arr[i] ;
   break;
  }
 }
}

Hi,firends we can also solve the above problem using Hash technique: Time Complexity:o(n) Space Complexity:o(n) Here is the code:

#include<bits/stdc++.h>
using namespace std;
int main(){
 int i,j,n;
 cout << "Enter the size of array:\n";
 cin >> n;

 int arr[n];
 //note that there is only one element which occure odd number of times
 cout << "Enter the elements of array:\n";
 for(i=0;i<n;i++){
  cin >> arr[i];
 }

 //using hash technique
 map<int,int> hash;
 for(i=0;i<n;i++){
  hash[arr[i]]++;
 }

 map<int,int>::iterator it;
 for(it=hash.begin();it!=hash.end();it++){
  if((it->second)%2==1){
   cout << it->first << endl;
   break;
  }
 }
}

Hi,guys but this above method is not optimise .

Concept:
The approach for this method is using XOR operation.

.XOR truth table:
1 XOR 1=0
0 XOR 0=0
1 XOR 0=1
0 XOR 1=1

.from xor truth table we can say that xor of a number the same number give 0 as result.

example:
4 XOR 4=0

Conclusion ::
if we take xor of a number with itself even number of times it gives 0.
but if we take xor of a number with itself odd number of times it gives number itself.

Here is the below code.still u have any doubt plz comment below.
Time Complexity:o(n)
Space Complexity:o(1)

#include<bits/stdc++.h>
using namespace std;
int main(){
 int i,j,n;
 cout << "Enter the size of array:\n";
 cin >> n;

 int arr[n];
 //note that there is only one element which occure odd number of times
 cout << "Enter the elements of array:\n";
 for(i=0;i<n;i++){
  cin >> arr[i];
 }

 //using XOR operation
 int ans;
 ans=arr[0];
 for(i=1;i<n;i++){
  ans=ans^arr[i];
 }

 cout << ans << endl;
}