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Reservoir Sampling


Reservoir Sampling is a technique of selecting k reservoir items randomly from a given list of n items, where n is very large.
For example, search lists in Google, YouTube etc.

Reservoir Sampling

Naive Approach for Reservoir Sampling

Build a reservoir array of size k, randomly select items from the given list. If the chosen item does not exist in the reservoir, add it, else continue for the next item.

  1. Create a reservoir array of size k.
  2. Initialize curr as 0, curr represent the curr position of reservoir array to be filled. Repeat step 3 and 4 while curr is less than k.
  3. Generate a random number between 0 to n(not included). Let the generated random number be rand.
  4. If the element at stream[rand] is already present in the reservoir array, repeat step 3. Else set reservoir[curr] as stream[rand] and increment curr.
  5. Print the elements of the reservoir array.

JAVA Code for Reservoir Sampling

public class ReservoirSampling {
    private static void selectReservoir(int[] stream, int k) {
        int n = stream.length;
        int reservoir[] = new int[k];
        // Index where the current item has to be placed
        int curr = 0;

        while (curr < k) {
            // Randomly generate an index between 0 to n
            int i = (int) (Math.random() * n) % n;
            boolean found = false;
            // Check if it is already present
            for (int j = 0; j < curr; j++) {
                if (reservoir[j] == stream[i]) {
                    found = true;
                    break;
                }
            }
            // If not present add it to the reservoir list
            if (!found) {
                reservoir[curr] = stream[i];
                curr++;
            }
        }

        // Print the randomly selected items
        for (int i = 0; i < k; i++) {
            System.out.print(reservoir[i] + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        // Example
        int stream[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
        int k = 5;
        selectReservoir(stream, k);
    }
}

C++ Code for Reservoir Sampling

#include <bits/stdc++.h>
using namespace std;

void selectReservoir(int stream[], int k, int n) {
    int reservoir[k];
    // Index where the current item has to be placed
    int curr = 0;
    
    srand(time(0));
    
    while (curr < k) {
        // Randomly generate an index between 0 to n
        int i = rand() % n;
        bool found = false;
        // Check if it is already present
        for (int j = 0; j < curr; j++) {
            if (reservoir[j] == stream[i]) {
                found = true;
                break;
            }
        }
        // If not present add it to the reservoir list
        if (!found) {
            reservoir[curr] = stream[i];
            curr++;
        }
    }
    
    // Print the randomly selected items
    for (int i = 0; i < k; i++) {
        cout<<reservoir[i]<<" ";
    }
    cout<<endl;
}

int main() {
    // Example
    int stream[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
    int k = 5;
    int n = sizeof(stream) / sizeof(stream[0]);
    selectReservoir(stream, k, n);
    return 0;
}
2 12 9 3 5

Output may differ, as the code chooses reservoirs randomly.

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Complexity Analysis

Time Complexity = O(k2
Space Complexity = O(1) 
where k is the number of elements present in the reservoir array.

Optimal Approach for Reservoir Sampling

The above problem can be solved efficiently for a given stream as,

  1. Create a reservoir array of size k and copy the first k items of stream into the array.
  2. Traverse the stream from index (k + 1) to n.
  3. For ith element in the stream generate a random number, say j, if the number lies in between 0 and (k – 1), replace reservoir[j] with stream[i].
  4. Reservoir array contains k random numbers selected from the stream, print them.

JAVA Code for Reservoir Sampling

public class ReservoirSampling {
    private static void selectReservoir(int[] stream, int k) {
        int n = stream.length;
        int reservoir[] = new int[k];

        // Copy first k items to the reservoir array
        for (int i = 0; i < k; i++) {
            reservoir[i] = stream[i];
        }

        // Traverse the remaining stream
        for (int i = k; i < n; i++) {
            // Generate a random number between 0 to n
            int randomNumber = (int) (Math.random() * n) % n;
            // If this number lies between 0 to k, replace reservoir[randomNumber] with stream of i
            if (randomNumber >= 0 && randomNumber < k) {
                reservoir[randomNumber] = stream[i];
            }
        }

        // Print the reservoir array
        for (int i = 0; i < k; i++) {
            System.out.print(reservoir[i] + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        int stream[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
        int k = 5;
        selectReservoir(stream, k);
    }
}

C++ Code for Reservoir Sampling

#include <bits/stdc++.h>
using namespace std;

void selectReservoir(int stream[], int k, int n) {
    int reservoir[k];
    
    // Copy first k items to the reservoir array
    for (int i = 0; i < k; i++) {
        reservoir[i] = stream[i];
    }
    
    srand(time(0));
    
    // Traverse the remaining stream
    for (int i = k; i < n; i++) {
        // Generate a random number between 0 to n
        int randomNumber = rand() % n;
        
        // If this number lies between 0 to k, replace reservoir[randomNumber] with stream of i
        if (randomNumber >= 0 && randomNumber < k) {
            reservoir[randomNumber] = stream[i];
        }
    }
    
    // Print the randomly selected items
    for (int i = 0; i < k; i++) {
        cout<<reservoir[i]<<" ";
    }
    cout<<endl;
}

int main() {
    // Example
    int stream[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
    int k = 5;
    int n = sizeof(stream) / sizeof(stream[0]);
    selectReservoir(stream, k, n);
    return 0;
}
11 2 8 4 6

Output may differ, as the code chooses reservoirs randomly.

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Complexity Analysis

Time Complexity = O(n) 
Space Complexity = O(1)
where n is the total number of elements in the stream and k is the number of elements present in the reservoir array.

References

Array Interview Questions
Graph Interview Questions
LinkedList Interview Questions
String Interview Questions
Tree Interview Questions
Core Java Interview Questions

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Abhishek was able to crack Microsoft after practicing questions from TutorialCup