Cracking the Coding Interview - Fooo

Solutions to Chapter 20 | Hard
20 2 Write a method to shuffle a deck of cards It must be a perfect shuffle - in other words,
each 52! permutations of the deck has to be equally likely Assume that you are given
a random number generator which is perfect
SOLUTION
2 8 1
Cracking the Coding Interview | Additional Review Problems
pg 91
This is a very well known interview question, and a well known algorithm If you aren’t one
of the lucky few to have already know this algorithm, read on
Let’s start with a brute force approach: we could randomly selecting items and put them into
a new array We must make sure that we don’t pick the same item twice though by somehow
marking the node as dead
Array: [1] [2] [3] [4] [5]
Randomly select 4: [4] [?] [?] [?] [?]
Mark element as dead: [1] [2] [3] [X] [5]
The tricky part is, how do we mark [4] as dead such that we prevent that element from being
picked again? One way to do it is to swap the now-dead [4] with the first element in the
array:
Array: [1] [2] [3] [4] [5]
Randomly select 4: [4] [?] [?] [?] [?]
Swap dead element: [X] [2] [3] [1] [5]
Array: [X] [2] [3] [1] [5]
Randomly select 3: [4] [3] [?] [?] [?]
Swap dead element: [X] [X] [2] [1] [5]
By doing it this way, it’s much easier for the algorithm to “know” that the first k elements are
dead than that the third, fourth, nineth, etc elements are dead We can also optimize this by
merging the shuffled array and the original array
Randomly select 4: [4] [2] [3] [1] [5]
Randomly select 3: [4] [3] [2] [1] [5]
This is an easy algorithm to implement iteratively:
1 public static void shuffleArray(int[] cards) {
2 int temp, index;
3 for (int i = 0; i < cards.length; i++){
4 index = (int) (Math.random() * (cards.length - i)) + i;
5 temp = cards[i];
6 cards[i] = cards[index];
7 cards[index] = temp;
8 }
9 }