Question

link

Given an array, generate a random permutation of array elements.

Solution

O(n) time complexity. 

To shuffle an array a of n elements (indices 0..n-1):
  for i from n − 1 downto 1 do
       j ← random integer with 0 ≤ j ≤ i
       exchange a[j] and a[i]

Note the RNG is having limit from 0 to i, and number i keeps decreasing.

Proof

This is called Fisher–Yates shuffle. Proof can be seen at question post:

The probability that ith element goes to second last position can be proved to be 1/n by dividing it in two cases.

Case 1: i = n-1 (index of last element):

The probability of last element going to second last position is = (probability that last element doesn’t stay at its original position) x (probability that the index picked in previous step is picked again so that the last element is swapped)

So the probability = ((n-1)/n) x (1/(n-1)) = 1/n

Case 2: 0 < i < n-1 (index of non-last):

The probability of ith element going to second position = (probability that ith element is not picked in previous iteration) x (probability that ith element is picked in this iteration)

So the probability = ((n-1)/n) x (1/(n-1)) = 1/n

We can easily generalize above proof for any other position.

Updated on Sep 10th, 2014: analysis of the approach. This question is on CC150v4 Q20.2.

Note that when we generate a new number between 0 and i, we swap it (with the last ‘alive’ number (ith number). After this, ith number is ‘dead’.

By doing it this way, we get a perfect shuffle! Idea is from cc150.

Updated again on Oct 2nd, 2014: I re-wrote the code for CC150v5 Q18.2. It’s very important to note that: 

if (a == b) {
    return;
}

When a == b, do not swap, otherwise the XOR swap method will product an zero!

Code

code form G4G, link

def sattoloCycle(items):
    i = len(items)
    while i > 1:
        i = i - 1
        j = randrange(i)  # 0 <= j <= i-1
        items[j], items[i] = items[i], items[j]
    return

written by me

public static void shuffleArrayInteratively(int[] cards) {
    for (int i = 0; i < cards.length; i++) {
        // all nums to the left of (i) is 'dead', don't consider them
        int choose = rand(i, cards.length - 1);
        swap(cards, i, choose);
        // now (i) is also 'dead'
    }
}

private static int rand(int from, int to) {
    int count = to - from + 1;
    return from + (int) (Math.random() * count);
}

private static void swap(int[] nums, int a, int b) {
    if (a == b) {
        return;
    }
    nums[a] ^= nums[b];
    nums[b] ^= nums[a];
    nums[a] ^= nums[b];
}