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Write a method for doing an in-place shuffle of an array.
**

The shuffle must be "uniform," meaning each item in the original array must have the same probability of ending up in each spot in the final array.

Assume that you have a method getRandom(floor, ceiling) for getting a random integer that is >= floor and <= ceiling.

A common first idea is to walk through the array and swap each element with a random other element. Like so:

import java.util.Random;
private static Random rand = new Random();
private static int getRandom(int floor, int ceiling) {
return rand.nextInt((ceiling - floor) + 1) + floor;
}
public static void naiveShuffle(int[] theArray) {
// for each index in the array
for (int firstIndex = 0; firstIndex < theArray.length; firstIndex++) {
// grab a random other index
int secondIndex = getRandom(0, theArray.length - 1);
// and swap the values
if (secondIndex != firstIndex) {
int temp = theArray[firstIndex];
theArray[firstIndex] = theArray[secondIndex];
theArray[secondIndex] = temp;
}
}
}

**However, this does not give a uniform random distribution.**

Why? We could calculate the exact probabilities of two outcomes to show they aren't the same. But the math gets a little messy. Instead, think of it this way:

Suppose our array had 3 elements: [a, b, c]. This means it'll make 3 calls to getRandom(0, 2). That's 3 random choices, each with 3 possibilities. So our total number of possible *sets of choices* is 3*3*3=27. Each of these 27 sets of choices is equally probable.

But how many possible *outcomes* do we have? If you paid attention in stats class you might know the answer is 3!, which is 6. Or you can just list them by hand and count:

a, b, c
a, c, b
b, a, c
b, c, a
c, b, a
c, a, b

But our method has 27 equally-probable sets of choices. 27 is not evenly divisible by 6. So some of our 6 possible *outcomes* will be achievable with more *sets of choices* than others.

We can do this in a single pass. time and space.

A common mistake is to have a mostly-uniform shuffle where an item is less likely to stay where it started than it is to end up in any given slot. Each item should have the same probability of ending up in each spot, including the spot where it starts.

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time and space.

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