Suppose we are given a $n$-vector $A$ of real numbers. We need to find the longest sequence of symmetrical numbers. The numbers $a_i$ and $a_j$ are symmetrical if they are located at the same distance from element $a_k$ and $|a_i-a_j|< \epsilon$, $\forall i < j$.


Can it be done in less than $O(n^2)$ comparisons in the worst case?

A $O(n^2)$ algorithm is to check the condition $|a_i-a_j|< \epsilon$ for element $a_k$.


Input: 0.006, 0.0061, 0.0069, 0.0135, 0.0065, 0.0193, 0.0086, 0.005, 0.01, 0.035, 0.065, 0.085, 0.1236, 0.086, 0.066, 0.037, 0.0024, 0.0712, 0.0032, 0.0174, 0.1504

There are two possible candidates of length 3 or more in this input (green and red).

1 (green). $a_k$ is 0.0065, and we have four pairs whose maximum absolute difference is 0.0058.

2 (red). $a_k$ is 0.1236, and we have three pairs whose maximum absolute difference is 0.002.


If $\epsilon > 0.0058$ then 0.035, 0.065, 0.085, 0.1236, 0.086, 0.066, 0.037

If $\epsilon <0.0058$ then 0.006, 0.0061, 0.0069, 0.0135, 0.0065, 0.0193, 0.0086, 0.005, 0.01

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Edit. After the comment by @D.W. It shound be noted that can be situation either $a_i = a_k$ or $a_j=a_k$, i.e. there is no the element $a_k$.

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  • 1
    $\begingroup$ If the condition was $a_i=a_j$ (instead of $|a_i-a_j|<\epsilon$), then it could can be solved in $O(n \log n)$ time. Let $h(a_i,a_{i+1},\dots,a_j)$ be a rolling hash of $a_i,\dots,a_j$. After a $O(n)$-time precomputation, you can compute $h(a_i,\dots,a_j)$ or $h(a_j,\dots,a_i)$ in $O(1)$ time for any $i,j$. Then, for each $k$, use binary search on $t$ to find the largest $t$ such that there is a symmetric sequence of length $2t$ centered at $k$, i.e., s.t. $h(a_{k-1},\dots,a_{k-t})=h(a_{k+1},\dots,a_{k+t})$. Your problem seems harder. (But maybe with locality sensitive hashing...?) $\endgroup$
    – D.W.
    Dec 27, 2023 at 23:55
  • 1
    $\begingroup$ Shouldn’t the solution be the first nine numbers? $\endgroup$
    – gnasher729
    Dec 29, 2023 at 3:11
  • $\begingroup$ @gnasher729, you are right. I have edited the post. $\endgroup$
    – Nick
    Dec 29, 2023 at 9:15

1 Answer 1


The way your condition is written, it could be that almost all elements are considered equal. That will be hard to handle efficiently.

By comparing $x_i$ and $x_{i+128}$ for example you can find all k that could be in the middle of a sequence with 129 or more elements. With your typical data that might very much restrict the values k, or there might be none at all and you compare $x_i$ and $x_{i+64}$ and so on. (Adjust depending on the number of items).

In your example you would find that $x_i$ and $x_{i+8}$ are close enough only for i = 0 and then compare 1 vs 7, 2 vs 6 and 3 vs 5.


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