# Geeksforgeeks Solution For " The XOR Gate "

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Problem :- Construct an N input XOR Gate. An XOR Gate returns 1 if odd number of its inputs are 1, otherwise 0.

Input:
The first line of input takes the number of test cases, T. Then T test cases follow.Each test case consists of 2 lines. The first line of each test case takes the number of inputs to the XOR Gate, N. The second line of each test case takes N space separated integers denoting the inputs to the  XOR Gate. Note that the inputs can be either 1's or 0's.

Output:
For each test case on a new line print the output of the N input XOR Gate.

Solution :-

#include <bits/stdc++.h>
using namespace std;
int findOdd(int arr[], int n)
{
int res = 0, i;
for (i = 0; i < n; i++)
res ^= arr[i];
return res;
}

int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,sum,res=0;
cin>>n;
int arr[n];
for(i=0;i<n;i++)
cin>>arr[i];
printf ("%d\n", findOdd(arr, n));
}
return 0;

}

Output :-

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#### post written by: Ghanendra Yadav

Hi, I’m Ghanendra Yadav, SEO Expert, Professional Blogger, Programmer, and UI Developer. Get a Solution of More Than 500+ Programming Problems, and Practice All Programs in C, C++, and Java Languages. Get a Competitive Website Solution also Ie. Hackerrank Solutions and Geeksforgeeks Solutions. If You Are Interested to Learn a C Programming Language and You Don't Have Experience in Any Programming, You Should Start with a C Programming Language, Read: List of Format Specifiers in C.