Algorithm to find two repeated numbers in an array, without sorting

Question

There is an array of size n (numbers are between 0 and n - 3) and only 2 numbers are repeated. Elements are placed randomly in the array.

E.g. in {2, 3, 6, 1, 5, 4

Answer 1

You know that your Array contains every number from 0 to n-3 and the two repeating ones (p & q). For simplicity, lets ignore the 0-case for now.

You can calculate the sum and the product over the array, resulting in:

1 + 2 + ... + n-3 + p + q = p + q + (n-3)(n-2)/2

So if you substract (n-3)(n-2)/2 from the sum of the whole array, you get

sum(Array) - (n-3)(n-2)/2 = x = p + q

Now do the same for the product:

1 * 2 * ... * n - 3 * p * q = (n - 3)! * p * q

prod(Array) / (n - 3)! = y = p * q

Your now got these terms:

x = p + q

y = p * q

=> y(p + q) = x(p * q)

If you transform this term, you should be able to calculate p and q

Answer 2

check this out ... O(n) time and O(1) space complexity

 for(i=0;i< n;i++)
 xor=xor^arr[i]
 for(i=1;i<=n-3;i++)
 xor=xor^i;

So in the given example you will get the xor of 3 and 5

xor=xor & -xor  //Isolate the last digit

for(i = 0; i < n; i++)
{
if(arr[i] & xor)
  x = x ^ arr[i]; 
else
  y = y ^ arr[i]; 
}
for(i = 1; i <= n-3; i++)
{
if(i & xor)
  x = x ^ i; 
else
  y = y ^ i; 

}

x and y are your answers

Answer 3

For each number: check if it exists in the rest of the array.

Answer 4

There is a O(n) solution if you know what the possible domain of input is. For example if your input array contains numbers between 0 to 100, consider the following code.

bool flags[100];
for(int i = 0; i < 100; i++)
    flags[i] = false;

for(int i = 0; i < input_size; i++)
    if(flags[input_array[i]])
         return input_array[i];
    else       
        flags[input_array[i]] = true;

Of course there is the additional memory but this is the fastest.

Answer 5

You might be able to take advantage of the fact that sum(array) = (n-2)*(n-3)/2 + two missing numbers.

Edit: As others have noted, combined with the sum-of-squares, you can use this, I was just a little slow in figuring it out.

Answer 6

Without sorting you're going to have a keep track of numbers you've already visited.

in psuedocode this would basically be (done this way so I'm not just giving you the answer):

for each number in the list
   if number not already in unique numbers list
      add it to the unique numbers list
   else
      return that number as it is a duplicate
   end if
end for each