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.

**My initial thoughts:**

We first select a card randomly from the 52 cards. That is probability of 1/52. We then select another card from the rest 51 cards. That is probability of 1/51. So on and so forth. Finally we have a particular permutation with probability of 1/52!, which is expected.

**My initial codes:**

public static List<Integer> shuffle() {
Set<Integer> deck = new HashSet<Integer>();
// build a deck of cards
for (int i = 0; i < 52; ++i) {
deck.add(i + 1);
}
List<Integer> shuffledDeck = new ArrayList<Integer>();
for (int i = 0; i < 52; ++i) {
int card;
do {
card = new Random().nextInt(52) + 1;
} while (!deck.contains(card));
deck.remove(card);
shuffledDeck.add(card);
}
return shuffledDeck;
}

**Solution:**

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:

public static void shuffleArray(int[] cards) {
int temp, index;
for (int i = 0; i < cards.length; i++) {
index = (int) (Math.random() * (cards.length - i)) + i;
temp = cards[i];
cards[i] = cards[index];
cards[index] = temp;
}
}

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Gayle Laakmann McDowell

May 29, 2012@ 01:19:38Hi Runhe,

I’ve been following your blog for a few months now, and I’ve noticed that while some of the posts are just your attempts at solving the problems, others actually re-type the book’s solutions.

Could you please:

(1) Remove the Cracking the Coding Interview tag on the posts

(2) Refrain from re-posting the book’s solution (which is in fact a violation of copyright law).

Thanks!

Gayle