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 a 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.
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 the probability of getting the last element as the result is 1/n.
Let us now talk about the second last element. When the second last element processed the first time, the probability that it replaced the previous result is 1/(n-1). The probability that the previous result stays when the nth item is considered is (n-1)/n. So the probability that the second last element is picked in the last iteration is [1/(n-1)] * [(n-1)/n] which is 1/n.
Similarly, we can prove for third last element and others.
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