## Three Interview Questions

### August 23, 2013

The first exercise is simple recursion through the tree:

`(define (bst? lt? t)`

(or (nil? t)

(and (or (nil? (lkid t))

(and (lt? (valu (lkid t)) (valu t))

(bst? (lkid t))))

(or (nil? (rkid t))

(and (lt? (valu t) (valu (rkid t)))

(bst? (rkid t)))))))

The second exercise isn’t much harder. We use a hash table, insert new items only if they don’t already exist, and return the result when we are finished. We could equally use a set:

`(define (dedup xs)`

(let ((h (make-hash)))

(do ((xs xs (cdr xs)))

((null? xs) (enlist h))

(when (null? (h 'lookup (car xs)))

(h 'insert (car xs))))))

The third exercise takes a little bit of thought. After the first pass over the fingers, the pattern repeats with a cycle of eight:

`(define (finger n)`

(case n

((1) 'thumb)

((2) 'index)

((3) 'middle)

((4) 'ring)

((5) 'pinkie)

(else (case (modulo (- n 5) 8)

((3) 'thumb)

((2 4) 'index)

((1 5) 'middle)

((0 6) 'ring)

((7) 'pinkie)))))

We used hash tables from a previous exercise (I’ve been too busy — or lazy — to put them in the Standard Prelude). You can run the program and see examples at http://programmingpraxis.codepad.org/ebAifNzO.

Pages: 1 2

finger seems a bit more complicated than it needs to be:

(define (finger n)

(case (modulo (- n 1) 8)

((0) ‘thumb)

((1 7) ‘index)

((2 6) ‘middle)

((3 5) ‘ring)

((4) ‘pinkie)))

Ooops – formatting.

Why not?

I concur with bhrgunatha. Indeed, it can be much simpler as he pointed out. The “minus one” can also be removed:

(define (fingersimpler n)

(case (modulo n 8)

((1) ‘thumb)

((2 0) ‘index)

((3 7) ‘middle)

((4 6) ‘ring)

((5) ‘pinkie)))

Also, in the original solution above, the calculation appears to be off by one for numbers greater than 5. For example, (finger 6) would return “middle”

About ‘remove duplicates from list’ (Python)

def remove_duplicates (l1):

return list (set (l1))

In Python:

I do not think that the solution presented for the BST check is correct.

From what I can make of the Scheme is that it only verifies that the immediate child nodes hold the BST property, but not ALL the descendants.

I think this is a common oversight, which makes it such a good interview question! :)

Let me present a solution that ensures for each node ALL right descendants are larger, and ALL left descendants are smaller or equal:

public class BinaryNode where T : IComparable {

public T Data { get; set; }

public BinaryNode Left { get; set; }

public BinaryNode Right { get; set; }

public bool IsBST() {

T max, min;

return DoBSTCheck (out max, out min);

}

private bool DoBSTCheck(out T max, out T min) {

max = Data;

min = Data;

if (Left != null) {

T lmax, lmin;

if (!Left.DoBSTCheck (out lmax, out lmin))

return false;

if (lmax.CompareTo(Data) > 0)

return false;

if (lmax.CompareTo(max) > 0)

max = lmax;

if (lmin.CompareTo(min) < 0)

min = lmin;

}

if (Right != null) {

T rmax, rmin;

if (!Right.DoBSTCheck (out rmax, out rmin))

return false;

if (rmin.CompareTo(Data) 0)

max = rmax;

if (rmin.CompareTo(min) < 0)

min = rmin;

}

return true;

}

}

[/sourceccode]

It is not terribly elegant (partly due to C#'s ugly generics code!), but the tracking of the min and max throughout the traversal does give an efficient solution!

Opps – bad code formatting. Fixed formatting: