I was reading an article on Longest Arithmetic Progression. The solution given has S(n)=$O(n^2)$. Can't I solve it in $O(1)$ space? To find the three elements, we first fix an element as middle element and search for other two (one smaller and one greater). When we found them we still keep on decreasing left and right to find llap (length of longest arithmetic progression). The other cases left are those where llap is even.. for them I will select i and (i+1)$^{th}$ element as my center two elements of AP and find elements on their right and left satisfying ap. T(n) will still be $O(n^2)$.

Is my approach correct?

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    $\begingroup$ This is the simple solution mentioned at the top of your link, which runs in cubic time. $\endgroup$ Aug 1, 2018 at 20:37
  • $\begingroup$ Why do you think the running time will be $O(n^2)$? In any case, many are often reluctant to check solutions here. If you're unsure, your best bet is to try to prove your answer; or you can ask about some specific aspect you're unsure about. $\endgroup$
    – D.W.
    Aug 1, 2018 at 21:11

1 Answer 1


No, your approach is wrong. Consider the set of numbers $1, 3, 4, 6, 7, 8, 10$. The longest arithmetic progression(LAP) in it is $1, 4, 7, 10$, which is of even length. However, 4 and 7 are not adjacent items so your approach will not find that LAP.

Yes, your approach is correct, but to a different problem from the problem in the article you mentioned.


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