Repeated and missing number in an array using xor - c

Question

How to find the repeated number and missing number as well using xor?
For eg: actual = [1,2,3] input_received = [3,2,3]. Here the missing number is 1 and the repeated number is 3. I found a quite interesting solution while surfing,
int missing_and_repeating(int a[], int n, int size){
int xor =0;
int i;
int x =0 , y =0;
for(i=0; i<size; i++)
xor = xor^a[i];
for(i=1; i<=n; i++)
xor = xor^i;
// Get the rightmost bit which is set
int set_bit_no = xor & ~(xor -1);
// XOR numbers in two buckets
for(i=0; i<size; i++){
if(a[i]& set_bit_no){
x = x^a[i];
}
else
y = y^ a[i];
}
for(i=1; i<=n; i++){
if(i & set_bit_no)
x = x^i;
else
y = y^i;
}
printf("\n %d %d ", x,y );
}
'actual' array is XORed and 'input_received' array is XORed
set_bit_no is assigned and both the arrays are split into two halves according to set_bit_no.
So again go back to our array and numbers from 1 to N-1 and 0 to size and XOR numbers in two buckets, one buckets contains XOR result of XORing all numbers with given bit set and other bucket contains XOR result of XORing all numbers with given bit reset.
I could not understand what set_bit_no is and why they are taking it, and how the array is split according to it. Someone please help me with a short example.

Answer 1

I assume that pre-condition is that exactly one array element is missing and instead replaced by a duplicate of another element. So, actual = {a1, a2, ..., an} and input_received = {a2, a2,..., an} (we can always rearrange elements so that a1 is missing and a2 is repeated).
So when you xor all elements from both arrays you get xored = a1 ^ a2.
Now we have to decompose that number to know a1 and a2. You take one of the non-zero bits (doesn't mater which one, the easiest way is to take the least significant like it's done in the code, and you always have one if a1 != a2). This bit is set in only one of numbers a1 or a2. So you xor all numbers which have this bit, and they'll annihilate each other, leaving only a1 or a2, and you xor all other numbers, so the result will be another number - a2 or a1 respectively.
It's important to note that this algorithm doesn't tell which of these numbers is missing and which is repeated. This is demonstrated by the following code:
int main() {
int arr1[] = {1, 1};
missing_and_repeating(arr1, 2, 2);
int arr2[] = {2, 2};
missing_and_repeating(arr2, 2, 2);
}
Output:
1 2
1 2