I have an array $a$ with $n$ elements, all of which have an associated weight. For example:

$a = \{ (A,2), (B,5), (C,9), ..., (Z,1) \}$, such that element $A$ has weight $w_A=2$, element $B$ has weight $w_B=5$ and so on. The array is not necessarily sorted.

Now, I would like to obtain a sample from this array where each element is selected without repetition and the probability is proportional to the weight of the element.

My goal is to sample approximately $k < n$ elements. I am emphasizing approximately because it does not have to be exactly $k$ but something close to $k$ is good enough. More precisely, I don't care if there is a deviation in the size of the final sample of $\pm \delta$ so long as $\delta$ is not too large in comparison to $k$ or $n$.

I know how to do the sampling in time $O(n + k \log n)$ by computing the cumulative weight $W$, then generating a random value in $[0,W]$ followed by a binary search to check which element this corresponds to.

However, my goal here is to understand whether it is possible to make a sampling in $O(n)$ by defining a probability $p_i$ for each element in such a way that I can decide independently for each element whether it should belong to the sample. Meanwhile, in the end I would still have a sample with approximately the size $k$ that I am looking for. If this is possible, then I could make a single pass in array $a$ after defining the probability $p_i$ accordingly which would require the procedure to define $p_i$'s to be $O(n)$ too, of course.

In summary, my question is: is there a way of defining these probabilities $p_i$ to achieve my goal? Or am I looking for an impossible solution here?

  • $\begingroup$ Can we assume that the elements are distinct? $\endgroup$
    – D.W.
    Jun 18, 2022 at 22:43
  • $\begingroup$ Why do you think that the method you proposed is correct? How does it ensure the "without repetition" part of the problem? $\endgroup$
    – D.W.
    Jun 20, 2022 at 0:06
  • 1
    $\begingroup$ Apparently as interpreted by me, if "each element is selected without repetition", then it cannot be true "the probability is proportional to the weight of the element". $\endgroup$
    – John L.
    Jun 23, 2022 at 21:39
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    $\begingroup$ If $k$ does not depend on $n$, $O(n+k\log n)$ is $O(n)$, so why worry ? $\endgroup$
    – user16034
    Jul 19, 2022 at 9:50

2 Answers 2


A solution should be like:

Assign the probability $p_i$ to the element $i$ equal to the normalized weight, $\dfrac{w_i}{w}$. Then perform $n$ systematic drawings that accept the element $i$ with probability $p_i=kw_i$, in turn. Then the expectation of the number of elements drawn is $k$ and the distribution is honored.

But this fails whenever $w_i>\dfrac1k$, which is no surprise because the elements may not be repeated, which limits their probability of appearance.


Yes, assuming we can assume that all elements are distinct. Here is one approach. Let $x_1,\dots,x_n$ denote the elements and $w_i$ the weight associated to $x_i$. Consider the following algorithm:

  • For $i:=1,2,\dots,n$:
    • Let $p_i=w_i/(w_i+w_{i+1}+\dots+w_n)$.
    • Draw $m_i$ from a Binomial($k$, $p_i$) distribution.
    • Output $m_i$ copies of $x_i$.
    • Set $k := k - m_i$.

This outputs the right distribution.

What is its running time? Well, you can compute the $p_i$'s in linear time, with a single backward scan. Also, you can sample a value $m$ from a Binomial($k$, $p$) distribution in $O(m)$ time. (Draw a uniform random number $q \in [0,1]$, then compute the tail sums of the Binomial, namely $\Pr[m\ge j]$ for $j:=0,1,2,\dots$, until you find the first tail sum that is $\ge q$. It takes a little bit of care to compute the tails sums, but you can do them in $O(1)$ time per tail sum.) Also, we have $m_1+\dots+m_n=k=O(n)$, so it follows that the total running time of this algorithm is $O(n)$.

I don't know whether this will be faster in practice. The constant factors might well dominate the difference between $O(1)$ vs $O(\log n)$.

  • $\begingroup$ All elements in the input array are distinct. However, the ~k elements in the sample should also be distinct. The third step says "output m_i copies of x_i" which would violate this characteristic, right? Or did I misunderstood your approach? And oh my, this got a lot more "complex" then I expected. You are right that it may not pay off in the end. $\endgroup$
    – Maltus
    Jun 19, 2022 at 1:32
  • $\begingroup$ @Maltus, sorry, I missed the "without repetition" part. Can you edit the question to clarify that all elements are distinct? $\endgroup$
    – D.W.
    Jun 20, 2022 at 0:06

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