Reservoir sampling is a family of randomized algorithms for randomly choosing k samples from a list of n items, where n is either a very large or unknown number. Typically n is large enough that the list doesn’t fit into main memory. For example, a list of search queries in Google and Facebook.
So we are given a big array (or stream) of numbers (to simplify), and we need to write an efficient function to randomly select k numbers where 1 <= k <= n. Let the input array be stream[].
A simple solution is to create an array reservoir[] of maximum size k. One by one randomly select an item from stream[0..n-1]. If the selected item is not previously selected, then put it in reservoir[]. To check if an item is previously selected or not, we need to search the item in reservoir[]. The time complexity of this algorithm will be O(k^2). This can be costly if k is big. Also, this is not efficient if the input is in the form of a stream.
It can be solved in O(n) time. The solution also suits well for input in the form of stream. The idea is similar to this post. Following are the steps.
1) Create an array reservoir[0..k-1] and copy first k items of stream[] to it.
2) Now one by one consider all items from (k+1)th item to nth item.
…a) Generate a random number from 0 to i where i is index of current item in stream[]. Let the generated random number is j.
…b) If j is in range 0 to k-1, replace reservoir[j] with arr[i]
void selectKItems(int stream[], int n, int k) { int i; // index for elements in stream[] // reservoir[] is the output array. Initialize // it with first k elements from stream[] int reservoir[k]; for (i=0;i<k;++i) reservoir[i]=stream[i]; // Use a different seed value so that we don't get // same result each time we run this program srand(time(NULL)); // Iterate from the (k+1)th element to nth element for (;i<n;++i) { // Pick a random index from 0 to i. int j=rand()%(i+1); // If the randomly picked index is in [0,k-1] // then replace the element present at the index // with new element from stream if (j<k) reservoir[j] = stream[i]; } cout << "Following are k randomly selected items \n"; for (int i=0;i<k;++i) cout<<reservoir[i]<<' '; } int main() { int stream[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; int n = sizeof(stream)/sizeof(stream[0]); int k = 5; selectKItems(stream, n, k); return 0; }
Time complexity of O (n)
Proof
The last k elements only the following two cases:
1. belonging steam [0 ~ k-1]
If an element belonging to the steam [0 ~ k-1], and finally also in the reservior, described later nk random elements are not randomized not to index the element.
P = (k/k+1)*(k+1/k+2)*...*(n-1/n) = k/n
2. belonging steam [k ~ n-1], the assumption is labeled element i, i∈ [k ~ n-1]
Description of the i-th element to the random [0 ~ k-1], and the subsequent elements there is no index to the random element.
P = (k/i+1)*(i+1/i+2)*...*(n-1/n) = k/n
=> The probability of each element is selected steam is k / n
Reference