Shuffle An Array
June 30, 2020
Today’s exercise comes to us from Leetcode via Reddit:
Given an array consisting of 2n elements in the form
[x1,x2,…,xn,y1,y2,…,yn], return the array in the form [x1,y1,x2,y2,…,xn,yn].
The Reddit poster claims to be new to Scheme and functional programming, and was thinking of a solution using length and list-ref, but couldn’t solve the problem.
Your task is to show the student how to solve the problem. 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.
Programming Praxis 
import random import itertools def shuffle(arr): """ """ l = len(arr) if l < 2 or l % 2 != 0: return [] l2 = l // 2 return list(itertools.chain(*[[arr[x], arr[x+l2]] for x in range(l2)])) n = 10 a = [random.randint(0, n * 2) for k in range(n)] print ("%s\n%s" % (a, shuffle(a)))Sample output:
Another version using generators
import random def shuffle2(arr): """ """ l = len(arr) if l < 2 or l % 2 != 0: return [] l2 = l // 2 for x in range(l2): yield arr[x] yield arr[x+l2] n = 10 a = [random.randint(0, n * 2) for k in range(n)] print ("%s\n%s" % (a, list(shuffle2(a))))A quick naive implementation in Haskell:
shuffle :: [Integer] -> [Integer] shuffle xs | length xs `mod` 2 == 0 = let n = div (length xs) 2 in foldr (\(x, y) acc -> x:y:acc) [] $ zip (take n xs) (drop n xs)Here’s a solution in Python.
def shuffle(l): return [x for pair in zip(l, l[len(l) // 2:]) for x in pair] print(shuffle(['1', '2', '3', '4', 'a', 'b', 'c', 'd']))Output:
A python implementation
def shuffle(xs): n = len(xs)//2 return list(reduce(λ a,b: a+b, zip(xs[:n], xs[n:])))A pythhon implementation
from functools import reduce def shuffle(xs): n = len(xs)//2 return list(reduce(lambda a,b: a+b, zip(xs[:n], xs[n:])))A C implementation
#include <stdlib.h> void *shuffle (int *arr, int len) { int n = len/2; int *first_half = (int *) calloc(n, sizeof(int)); int *second_half = (int *) calloc(n, sizeof(int)); int i, j; for (i=0; i < n; ++i) { first_half[i] = arr[i]; second_half[i] = arr[i+n]; } for (i=0; i<n; ++i) { j = 2*i; arr[j] = first_half[i]; arr[j+1] = second_half[i]; } free (first_half); free (second_half); }Clojure:
(defn shuffle-array [arr] (let [half (/ (count arr) 2)] (into [] (interleave (take half arr) (drop half arr)))))Test:
Though checks can be added to ensure that input is a sequence that has even amount of elements:
(defn shuffle-array [arr] (if (and (sequential? arr) (= 0 (mod (count arr) 2))) (let [half (/ (count arr) 2)] (into [] (interleave (take half arr) (drop half arr)))) []))Test: