Puzzle : finding out repeated element in an Array

Question

Size of an array is n.All elements in the array are distinct in the range of [0 , n-1] except two elements.Find out repeated element without using extra temporary array with con

Answer 1

O(n) without the temp array.

a[]={1,0,0,2,3};
i=0;
int required;
while(i<n)
{
  a[a[i] % n] += n;
  if(a[a[i] % n] >= 2 * n)
    required = a[i] % n;
}
print required;

(Assuming of course that n < MAX_INT - 2n)

Answer 2

This example could be useful for int, char, and string.

char[] ch = { 'A', 'B', 'C', 'D', 'F', 'A', 'B' };
Dictionary<char, int> result = new Dictionary<char, int>();
foreach (char c in ch)
{
   if (result.Keys.Contains(c))
   {
       result[c] = result[c] + 1;
   }
   else
   {
       result.Add(c, 1);
   }
}
foreach (KeyValuePair<char, int> pair in result)
{
   if (pair.Value > 1)
   {
       Console.WriteLine(pair.Key);
   }
}
Console.Read();

Answer 3

Based on @sje's answer. Worst case is 2 passes through the array, no additional storage, non destructive.

O(n) without the temp array.

a[]={1,0,0,2,3};
i=0;
int required;
while (a[a[i] % n] < n)    
   a[a[i++] % n] += n;

required = a[i] % n;
while (i-->0)
   a[a[i]%n]-=n;

print required;

(Assuming of course that n < MAX_INT/2)