Select a random number from stream, with O(1) space

Given a stream of numbers, generate a random number from the stream. You are allowed to use only O(1) space and the input is in the form of stream, so can’t store the previously seen numbers.

So how do we generate a random number from the whole stream such that the probability of picking any number is 1/n. with O(1) extra space? This problem is a variation of Reservoir Sampling. Here the value of k is 1.

1) Initialize ‘count’ as 0, ‘count’ is used to store count of numbers seen so far in stream.
2) For each number ‘x’ from stream, do following
…..a) Increment ‘count’ by 1.
…..b) If count is 1, set result as x, and return result.
…..c) Generate a random number from 0 to ‘count-1’. Let the generated random number be i.
…..d) If i is equal to ‘count – 1’, update the result as x.

C/C++

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// An efficient C program to randomly select a number from stream of numbers.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
  
// A function to randomly select a item from stream[0], stream[1], .. stream[i-1]
int selectRandom(int x)
{
    static int res;    // The resultant random number
    static int count = 0;  //Count of numbers visited so far in stream
  
    count++;  // increment count of numbers seen so far
  
    // If this is the first element from stream, return it
    if (count == 1)
        res = x;
    else
    {
        // Generate a random number from 0 to count - 1
        int i = rand() % count;
  
        // Replace the prev random number with new number with 1/count probability
        if (i == count - 1)
            res  = x;
    }
    return res;
}
  
// Driver program to test above function.
int main()
{
    int stream[] = {1, 2, 3, 4};
    int n = sizeof(stream)/sizeof(stream[0]);
  
    // Use a different seed value for every run.
    srand(time(NULL));
    for (int i = 0; i < n; ++i)
        printf("Random number from first %d numbers is %d \n",
                                i+1, selectRandom(stream[i]));
    return 0;
}

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Java

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//An efficient Java program to randomly select a number from stream of numbers.
  
import java.util.Random;
  
public class GFG 
{
    static int res = 0;    // The resultant random number
    static int count = 0//Count of numbers visited so far in stream
      
    //A method to randomly select a item from stream[0], stream[1], .. stream[i-1]
    static int selectRandom(int x)
    {
        count++; // increment count of numbers seen so far
          
        // If this is the first element from stream, return it
        if (count == 1)
            res = x;
        else
        {
             // Generate a random number from 0 to count - 1
            Random r = new Random();
            int i = r.nextInt(count);
              
            // Replace the prev random number with new number with 1/count probability
            if(i == count - 1)
                res = x;
        }
        return res;
    }
      
    // Driver program to test above function.
    public static void main(String[] args)
    {
        int stream[] = {1, 2, 3, 4};
        int n = stream.length;
        for(int i = 0; i < n; i++)
            System.out.println("Random number from first " + (i+1) +
                               " numbers is " + selectRandom(stream[i]));
    }
}
//This code is contributed by Sumit Ghosh

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Python3

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# An efficient pyhton3 program 
# to randomly select a number
# from stream of numbers.
import random
  
# A function to randomly select a item
# from stream[0], stream[1], .. stream[i-1]
def selectRandom(x):
      
    # The resultant random number
    res = 0;
      
    # Count of numbers visited 
    # so far in stream
    count = 0;
  
    # increment count of numbers 
    # seen so far
    count += 1;
  
    # If this is the first element 
    # from stream, return it
    if (count == 1):
        res = x;
    else:
          
        # Generate a random number 
        # from 0 to count - 1
        i = random.randrange(count);
  
        # Replace the prev random number 
        # with new number with 1/count 
        # probability
        if (i == count - 1):
            res = x;
    return res;
  
# Driver Code
stream = [1, 2, 3, 4];
n = len(stream);
  
# Use a different seed value 
# for every run.
for i in range (n):
    print("Random number from first"
         (i + 1), "numbers is"
          selectRandom(stream[i]));
  
# This code is contributed by mits

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C#

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// An efficient C# program to randomly 
// select a number from stream of numbers.
using System;
  
class GFG 
{
    // The resultant random number
    static int res = 0; 
      
    // Count of numbers visited
    // so far in stream
    static int count = 0; 
      
    // A method to randomly select 
    // a item from stream[0], 
    // stream[1], .. stream[i-1]
    static int selectRandom(int x)
    {
        // increment count of
        // numbers seen so far
        count++; 
          
        // If this is the first 
        // element from stream, 
        // return it
        if (count == 1)
            res = x;
        else
        {
            // Generate a random number 
            // from 0 to count - 1
            Random r = new Random();
            int i = r.Next(count);
              
            // Replace the prev random 
            // number with new number 
            // with 1/count probability
            if(i == count - 1)
                res = x;
        }
        return res;
    }
      
// Driver Code
public static void Main()
{
        int[] stream = {1, 2, 3, 4};
        int n = stream.Length;
        for(int i = 0; i < n; i++)
            Console.WriteLine("Random number from "
                              "first {0} numbers is {1}" ,
                          i + 1, selectRandom(stream[i]));
    }
}
  
// This code is contributed by mits

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PHP

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<?php
// An efficient php program to randomly 
// select a number from stream of numbers.
  
// A function to randomly select a item 
// from stream[0], stream[1], .. stream[i-1]
function selectRandom($x)
{
      
    // The resultant random number
    $res;
      
    // Count of numbers visited so far
    // in stream
    $count = 0;
  
    // increment count of numbers seen
    // so far
    $count++;
  
    // If this is the first element 
    // from stream, return it
    if ($count == 1)
        $res = $x;
    else
    {
          
        // Generate a random number from
        // 0 to count - 1
        $i = rand() % $count;
  
        // Replace the prev random number 
        // with new number with 1/count 
        // probability
        if (i == $count - 1)
            $res = $x;
    }
    return $res;
}
  
// Driver program to test above function.
    $stream = array(1, 2, 3, 4);
    $n = sizeof($stream)/sizeof($stream[0]);
  
    // Use a different seed value for
    // every run.
    srand(time(NULL));
      
    for ($i = 0; $i < $n; ++$i)
        echo "Random number from first "
                   $i+1, "numbers is " ,
        selectRandom($stream[$i]), "\n" ;
  
// This code is contributed by nitin mittal.
?>

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Output:

Random number from first 1 numbers is 1
Random number from first 2 numbers is 1
Random number from first 3 numbers is 3
Random number from first 4 numbers is 4

Auxiliary Space: O(1)



How does this work
We need to prove that every element is picked with 1/n probability where n is the number of items seen so far. For every new stream item x, we pick a random number from 0 to ‘count -1’, if the picked number is ‘count-1’, we replace the previous result with x.

To simplify proof, let us first consider the last element, the last element replaces the previously stored result with 1/n probability. So probability of getting last element as result is 1/n.

Let us now talk about second last element. When second last element processed first time, the probability that it replaced the previous result is 1/(n-1). The probability that previous result stays when nth item is considered is (n-1)/n. So probability that the second last element is picked in last iteration is [1/(n-1)] * [(n-1)/n] which is 1/n.

Similarly, we can prove for third last element and others.

References:
Reservoir Sampling

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : nitin mittal, Mithun Kumar