There is a permutation (more precisely a derangement) $\sigma$ of the set $\{0,1,\dots,n-1\}$ with cardinality $n$.

I want to compute the following counts (a kind of inversion):

$$K(\sigma )_{i}=\#\{j>i:\sigma _{j}>i\}$$

for each $0 \le i \lt n$.

Obviously a $O(n^2)$ straight-forward algorithm computes these counts. But can it be done faster (eg in $O(n \log n)$)?

I can't seem to wrap my head around such an algorithm, based on other divide-and-conquer algorihms for usual inversions, at least so far.

Background: The counts above are used in a custom algorithm to rank and unrank derangements in lexicographic order and their computation is the main bottleneck of the algorithm.

  • $\begingroup$ Visualising the problem helps here. You have got a permutation matrix, represented by recording for each row the column where it has a nonzero entry. In $O(n)$ time, you can get the "transposed" information (corresponding to $\sigma^{-1}$) as well, if you should need it, The problem is to count, for each $i$, the sum of entries in the bottom-right square $(n-i)\times(n-i)$ matrix; by looking at the differences for adjacent $i$, this can be done in a single pass. $\endgroup$ Commented Jul 11, 2021 at 11:39

1 Answer 1


Each element $j$ contributes $1$ to the cardinality of all sets $\{j > i \mid \sigma_j > i\}$ for which $i < \min\{\sigma_j, j\}$, and $0$ to the other sets.

You can compute all $n$ values $K(\sigma)_i$ in $O(n)$ time as follows. Maintain an array $A[0, \dots, n-1]$ where each entry $A[i]$ is initialized to $0$. Then, for each $j$, increment $A[\min\{\sigma_j, j\}]$ by $1$.

Compute the sums of the elements in all suffixes of $A$, i.e., construct a new array $K[1, \dots, n-1]$ such $K[i] = \sum_{i' > i} A[i']$. This can be done in $O(n)$ time by setting $K[n-1]=0$ and, for all $i=n-2, \dots, 0$ (in this order), $K[i] = K[i+1] + A[i+1]$.

Clearly, the above can also be done in-place without the need of the additional array $K$.

  • $\begingroup$ Excellent! Can a similar strategy be applied to count $K'(\sigma )_{i}=\#\{j<i:\sigma _{j}>i\}$?? $\endgroup$
    – Nikos M.
    Commented Jul 10, 2021 at 14:56
  • 1
    $\begingroup$ $$|\{ j < i : \sigma_j > i \}| + |\{ j > i : \sigma_j > i \}| + |\{ i : \sigma_i > i \}| = |\{ j \mid \sigma_j > i \}.|$$ Substituting: $$ K'(\sigma)_i + K(\sigma)_i + 1_{\sigma_i > i} = n-i-1. $$ Therefore: $$ K'(\sigma)_i = n-i-1 - K(\sigma)_i - 1_{\sigma_i > i} $$ $\endgroup$
    – Steven
    Commented Jul 10, 2021 at 15:02
  • $\begingroup$ hmm, $K'(\sigma)$ is needed in unranking, and we dont have $K(\sigma)$ as we dont have yet the permutation, it is built step-by-step $\endgroup$
    – Nikos M.
    Commented Jul 10, 2021 at 15:04
  • $\begingroup$ Uhm... $K'(\sigma)_i$ is defined as a function of sigma. How would you compute $K'(\sigma)_i$ in the first place (even in a non-efficient way) if you don't know $\sigma$? $\endgroup$
    – Steven
    Commented Jul 10, 2021 at 15:06
  • $\begingroup$ $K'(\sigma)$ is based strictly on previous entries $\endgroup$
    – Nikos M.
    Commented Jul 10, 2021 at 15:07

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