Determine if an array from 1..n has a duplicate in constant time and space.

The interviewer is asking for an algorithm to determine if an array with values ranging from 1 to n includes any duplicate values. There are two additional constraints that make this task more challenging.

The first constraint is to solve this problem in constant time complexity: regardless of the size of the array, the algorithm's performance should remain the same.

The second constraint is to solve this problem with constant space complexity: the amount of memory used to solve the problem should not depend on the size of the array.

To answer this question effectively, the key points to remember include:

1. The algorithm needs to run in constant time - this means loops, recursive calls, and any operation that involves checking each element are not effective solutions.
2. The algorithm also needs to run in constant space - this means creating a data structure to store elements or copies of the array is not an advisable solution.

For an array of size n with elements ranging from 1 to n, there should ideally be no duplicates because each element can be present exactly once.

That, however, is the ideal case. In real scenarios, this array can contain duplicates.

Here's a trick we can use to find duplicates: Since all the elements are positive, calculate the sum of all elements of the array and compare it with sum of first n natural numbers as given by the formula (n (n + 1)) / 2.

If the sum of elements of array is greater than the sum of first n natural numbers, we can conclude that there is a duplicate.

However this approach will not work if there are multiple duplicates, or numbers absent from the array. There is no straight forward solution to this question that would take O(1) time and O(1) space, due to restrictions of time and space complexity.

A different approach would be using mathematical properties, specifically properties of arithmetic series. You could calculate the expected sum of an arithmetic series using the formula n*(n+1)/2.

Next, subtract from it the actual sum of the elements in the array (which requires to sum over the array). If the array contains one duplicate and no number is missing in 1..n, the result is the duplicate number. If the result is zero, no duplicate exists.

It's important to note that this approach will work under the conditions that only one number is a duplicate and that no numbers are missing in the series from 1 to n. If these conditions are not met, it's impossible to solve this problem under the O(1) time and O(1) space constraints.

This solution is limited and doesn't cover all edge cases, and it's crucial to convey this to the interviewer.