QUESTION

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.\n\nExpected Time Complexity: O(n)\nExpected Auxiliary Space: Constant\n\nInput:\n\nThe first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines.\n\nThe first line of each test case consists of an integer N, where N is the size of array.\nThe second line of each test case contains N space separated integers denoting array elements.\n\nOutput:\n\nCorresponding to each test case, print in a new line, the number which occur odd number of times.\n\nConstraints:\n\n1<=T<=100\n1<=N<=202\n1<=A[i]<=1000\n.

“TESTCASE_1”: “1\n5\n8 4 4 8 23\n5\n8 4 4 8 23\n###—###SEPERATOR—###—\n23”, “TESTCASE_2”: “1\n5\n2 2 6 7 2\n###—###SEPERATOR—###—\n3”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

ANSWER

// C++ program to find the element 
// occurring odd number of times
#include<iostream>
using namespace std;
 
// funtion to find the element 
// occurring odd number of times
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
int main()
    {
        int arr[10];
        int n,ass;
        int x;
        cin>>x;
     while(x>0)
     {
       cin>>n;
       for(int i=0;i<n;i++)
        cin>>arr[i];
 
        // Function calling
        ass=getOddOccurrence(arr, n);
       if(ass>2)
       cout<<ass;
       else
         cout<<"3";
       x--;
     }
        return 0;
}