find all repeating in array O(1) space

Find all repeating numbers in array of size n with range 0..n-1. Numbers could repeat any number of times. In O(1) space and O(n) time. (related to this)

The range is 0 .. n-1 and array is of size n. The array itself is used as flags by mutating it. The range is not -ve so a -ve value is used as flag. For x, a[x] is set to -ve on 1st visit if +ve, and added to dupes if -ve (after 1st visit). Note that the logic deals with item x and value a[x] where x = abs(x). As a[x] can be -ve, abs is needed.

i:   i  ..   x 
a: a[i] .. a[x] //x = abs(a[i])

For x =  2, a[2] is set to -2 from 2.
For x = -2, a[2] is -ve so added to dupes.
i: 0  1  2
a: .  2 -2
         -

//example
     - -
a: 1 2 1
i: 0 1 2
dupes: 1

// does not work if a[i] = 0
(a):
    if a == null:
        throw
    set dupes = new
    for x in a:
        x = abs(x)
        if a[x] < 0:
            dupes.add(x)
        else: // a[x] >= 0
            a[x] = -a[x]
    return dupes

When a[x] is 0, set it to -n to be used as a flag (0 cannot be used as flag). Which means a[i] of -n denotes 0 so %n to handle it.

//example
  -n
a: 0 0 0
i: 0 1 2
dupes: 0

   --n
a: 1 0 0 1
i: 0 1 2 3
dupes: 0 1

// O(n) time, O(1) space
(a):
    if a == null:
        throw
    set dupes = new
    n = a.length()
    for x in a:
        x = abs(x) %n     // to handle 0 values
        if a[x] < 0:
            dupes.add(x)
        else if a[x] > 0:
            a[x] = -a[x]
        else: //a[x] == 0 // to handle 0 values
            a[x] = -n
    return dupes