Reputation: 17
XOR is working fine for single digit inputs but totally messes up while using 2 or more digits number. How can I still make this program by using XOR ?
This is the question basically :
Find the Missing Number
You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in list. One of the integers is missing in the list. Write an efficient code to find the missing integer.
My program is:
#include<iostream>
using namespace std;
int main(void)
{
int n; //number of elements
cout << "Enter the number of elements : ";
cin >> n;
//Declaring array
int arr[n-1];
//Taking input
for(int i = 0; i<n-1; i++)
cin >> arr[i];
//printing array
cout << "\nElements are :";
for(int i = 0; i<n-1; i++)
cout << " " << arr[i];
//Declaring elements to take XOR
int x1 = arr[0]; // first element of array
int x2 = 1; //first element in natural number series i.e 1,2,3,4...
//taking XOR of all elements of the given array
for(int i = 1; i<n-1; i++)
x1 ^= arr[i];
//taking XOR of all the natural numbers till 'n'
for(int i = 2; i<arr[n-1]; i++)
x2 ^= i;
//Finally printing the output by taking XOR of x1 and x2
//because same numbers will be removed since (A^A) = 0 also (A^0) = A
cout << "\nMissing number : " << (x1^x2) << endl;
return 0;
}
The above program doesn't work for the input below:
10
1 2 3 4 5 6 7 8 10
Upvotes: 0
Views: 259
Reputation: 217135
Your loops are wrong, you can change to:
//Declaring elements to take XOR
int x1 = 0; // for element of array
int x2 = 0; // for natural number series i.e 1,2,3,4...
//taking XOR of all elements of the given array
for (int i = 0; i < n-1; i++)
x1 ^= arr[i];
//taking XOR of all the natural numbers till 'n'
for (int i = 1; i != n + 1; i++)
x2 ^= i;
Note that natural number range is bigger than arr size.
Upvotes: 1