The Schensted insertion algorithm has an $O(n^2)$ running time, for constructing such a standard Young's Tableaux.

But, since every permutation has a unique Young's tableau, there seems no reason as to why it cannot be done in $O(n \log n)$ time, which is the optimal time to identify a permutation.

Are there any known non-trivial $\textbf{lower bounds or upper bounds}$ for this?

Any assistance/guidance towards any of the questions is much appreciated. Thanks in advance

Sample input permutation


Output Young's Tableaux


  • $\begingroup$ I suggest editing the question to incorporate that information, so the question stands on its own. We want questions to be self-contained, so people don't need to read the comments to understand what you are asking. Also, it sounds like the correspondence is to pairs of Young tableaux, not to a single tableux; if that is correct, please make the corresponding change as well. Thank you! $\endgroup$
    – D.W.
    Commented Feb 8, 2017 at 4:02
  • $\begingroup$ @D.W:- No Sir, i do not need pair of Young tableaux, just a single tableaux. For eg. the tableaux on page en.wikipedia.org/wiki/Robinson%E2%80%93Schensted_correspondence Corresponds to the permutation 8,3,1,5,2,4,7 $\endgroup$
    – Vk1
    Commented Feb 8, 2017 at 4:09
  • 1
    $\begingroup$ I think you need to specify the desired output more precisely. When you say that every permutation corresponds to a unique Young's tableaux, what correspondence do you have in mind? The correspondence listed on Wikipedia is between a permutation and a pair of standard Young's tableaux (not between a permutation and a single Young's tableaux). Perhaps given a permutation, you want the first of the pair of tableaux that correspond to it? $\endgroup$
    – D.W.
    Commented Feb 8, 2017 at 4:22
  • $\begingroup$ @D.W. :- I have made the changes, and given an example in the description. I hope this clarifies everything. $\endgroup$
    – Vk1
    Commented Feb 8, 2017 at 13:16

1 Answer 1


I believe that no faster algorithm is known, at least if you are interested in computing both tableaux. Dan Romik shows in his paper The Number of Steps in the Robinson-Schensted Algorithm that on almost all permutations, the computational complexity is actually only $\Theta(n^{3/2} \log n)$. Duzhin, Kuzmin and Vasilliev show in their work RSK bumping trees and a fast RSK algorithm how to speed up the algorithm in practice by a significant constant factor. This is all I could find on the topic.

  • $\begingroup$ I would hug you, if you were besides me right now. How did you know all this? You have no idea how much I am grateful to you. Thank you so much. $\endgroup$
    – Vk1
    Commented Jun 10, 2022 at 15:27
  • $\begingroup$ I just googled some keywords. $\endgroup$ Commented Jun 10, 2022 at 15:48

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