Here's a sketchy description for a possible solution.

The easy part is that we know that if x and y are the missing and duplicate indices form vector V then we have:

    |x-y| = |sum(V) - (N*(N-1))/2|

So far - constant space and linear time. From here it's either cleverly (linearly!) scanning the vector or similarly concluding something else about the vector and crossing that info with what we have ;)