## Swap list nodes

### June 25, 2013

We have today another from our never-ending list of interview questions:

Given a linked list, swap the

kth node from the head of the list with thekth node from the end of the list.

Your task is to write a function to perform the indicated task; be sure to test it thoroughly. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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;; 1- test cases:

(defun test/swap-kth ()

(loop

:for (expected list k) :in ‘((() () 0)

(() () 1)

((x) (x) 0)

((x) (x) 1)

((b a) (a b) 0)

((b a) (a b) 1)

((a b) (a b) 2)

((c b a) (a b c) 0)

((a b c) (a b c) 1)

((c b a) (a b c) 2)

((a b c) (a b c) 3)

((d b c a) (a b c d) 0)

((a c b d) (a b c d) 1)

((a c b d) (a b c d) 2)

((d b c a) (a b c d) 3)

((a b c d) (a b c d) 4)

)

:for mutable = (copy-list list)

:for result = (swap-kth mutable k)

:do (assert (equal expected mutable)

() "(~S ‘~S ~S) -> ~S instead of expected ~S"

‘swap-kth list k result expected))

:success)

;; 2- a correct implementation.

(defun swap-kth (list k)

(check-type k (integer 0))

(when (< k (length list))

(rotatef (nth k list) (nth (- (length list) k 1) list)))

list)

(test/swap-kth)

;; –> :success

;; Notice that this correct implementation is O(n) time and O(1)

;; space, so it’s not bad. But it walks the list up to 4 times (could

;; be 3 times, factorizing out (length list).

;; We can write an optimized solution that walks the list only up to 2

;; times with O(1) space, or only once with O(n) space.

;; 3- an optimized solution, with still O(1) space:

;; we walk the list once to compute the length, and then down

;; (max k (- length k 1)).

(defun swap-kth (list k)

(check-type k (integer 0))

(let ((len (length list))) ; 1st walk.

(when (< k len)

(let* ((l (- len k 1))

(i (min k l))

(j (max k l))

(a (nthcdr i list))

(b (nthcdr (- j i) a)))

(rotatef (car a) (car b))))

list))

(test/swap-kth)

;; –> :success

;; 4- an optimized solution using Θ(n) space, but amortized only Θ(n)

;; time, walking the list only once. It builds a vector parallel

;; to the list.

(defun swap-kth (list k)

(check-type k (integer 0))

(let ((v (make-array 1 :adjustable t :fill-pointer 0)))

(loop

:for cell :on list

:do (vector-push-extend cell v (array-dimension v 0)))

(let ((len (length v))) ; O(1)

(when (< k len)

(let* ((l (- len k 1)))

(rotatef (car (aref v k)) (car (aref v l))))))

list))

(test/swap-kth)

;; –> :success

Here’s a Rackety version (although this one should work in most any Scheme): Swap list nodes on jverkamp.com

Check out the link above for test cases and a bit of discussion. Two9 things that were particularly neat was the use of

`eq?`

for quick pointer comparison and the relative simplicity of the code so far as dealing with special cases goes. Granted, it doesn’t bail early if`kth-head`

is`eq?`

to`kth-tail`

, but that would be easy enough to add.Here’s my version. Not particularly elegant.

The real work is done in swap__ appending the results of a take, a drop-take, a drop, and two list-refs. The swap_ function just checks for edge cases. The body of the outer function checks the types of the parameters and allows the user to pass them in either order.

def change_list(mylist,place):

temp = mylist[place-1]

mylist[place-1] = mylist[len(mylist)-place]

mylist[len(mylist)-place] = temp

return mylist

`mylist = [1,2,3,4,5,6,7,8,9]`

print (change_list(mylist,6))