Algorithm to determine if array contains n…n+m?

Question

I saw this question on Reddit, and there were no positive solutions presented, and I thought it would be a perfect question to ask here. This was in a thread about interview

Answer 1

Counter-example for XOR algorithm.

(can't post it as a comment)

@popopome

For a = {0, 2, 7, 5,} it return true (means that a is a permutation of the range [0, 4) ), but it must return false in this case (a is obviously is not a permutaton of [0, 4) ).

Another counter example: {0, 0, 1, 3, 5, 6, 6} -- all values are in range but there are duplicates.

I could incorrectly implement popopome's idea (or tests), therefore here is the code:

bool isperm_popopome(int m; int a[m], int m, int  n)
{
  /** O(m) in time (single pass), O(1) in space,
      no restrictions on n,
      no overflow,
      a[] may be readonly
  */
  int even_xor = 0;
  int odd_xor  = 0;

  for (int i = 0; i < m; ++i)
    {
      if (a[i] % 2 == 0) // is even
        even_xor ^= a[i];
      else
        odd_xor ^= a[i];

      const int b = i + n;
      if (b % 2 == 0)    // is even
        even_xor ^= b;
      else
        odd_xor ^= b;
    }

  return (even_xor == 0) && (odd_xor == 0);
}