为什么这个算法的Big-O N ^ 2 * log N.

Question

Fill array a from a[0] to a[n-1]: generate random numbers until you get one that is not already in the previous indexes.

This is my implementation:

public static int[] first(int n) {
    int[] a = new int[n];
    int count = 0;

    while (count != n) {
        boolean isSame = false;
        int rand = r.nextInt(n) + 1;

        for (int i = 0; i < n; i++) {
            if(a[i] == rand) isSame = true;
        }

        if (isSame == false){
            a[count] = rand;
            count++;
        }
    }

    return a;
}

I thought it was N^2 but it's apparently N^2logN and I'm not sure when the log function is considered.

Answer 1

The 0 entry is filled immediately. The 1 entry has probability 1 - 1 / n = (n - 1) / n of getting filled by a random number. So we need on average n / (n - 1) random numbers to fill the second position. In general, for the k entry we need on average n / (n - k) random numbers and for each number we need k comparisons to check if it's unique.

So we need

n * 1 / (n - 1) + n * 2 / (n - 2) + ... + n * (n - 1) / 1

comparisons on average. If we consider the right half of the sum, we see that this half is greater than

n * (n / 2) * (1 / (n / 2) + 1 / (n / 2 - 1) + ... + 1 / 1)

The sum of the fractions is known to be Θ(log(n)) because it's an harmonic series . So the whole sum is Ω(n^2*log(n)) . In a similar way, we can show the sum to be O(n^2*log(n)) . This means on average we need

Θ(n^2*log(n))

operations.

Answer 2

This is similar to the Coupon Collector problem. You pick from n items until you get one you don't already have. On average, you have O(n log n) attempts (see the link, the analysis is not trivial). and in the worst case, you examine n elements on each of those attempts. This leads to an average complexity of O(N^2 log N)

Answer 3

The algorithm you have is not O(n^2 lg n) because the algorithm you have may loop forever and not finish. Imagine on your first pass, you get some value $X$ and on every subsequent pass, trying to get the second value, you continue to get $X$ forever. We're talking worst case here, after all. That would loop forever. So since your worst case is never finishing, you can't really analyze.

In case you're wondering, if you know that n is always both the size of the array and the upper bound of the values, you can simply do this:

int[] vals = new int[n];
for(int i = 0; i < n; i++) {
    vals[i] = i;
}
// fischer yates shuffle
for(int i = n-1; i > 0; i--) {
   int idx = rand.nextInt(i + 1);
   int t = vals[idx];
   vals[idx] = vals[i];
   vals[i] = t;
}

One loop down, one loop back. O(n) . Simple.

Answer 4

If I'm not mistaken, the log N part comes from this part:

for(int i = 0; i < count; i++){
    if(a[i] == rand) isSame = true;
}

Notice that I changed n for count because you know that you have only count elements in your array on each loop.