We are given two numbers N and K.

N <= 10^9.


We need to find numbers of permutations of ( 1 to N ) such that inversions are exactly K.

If N was <= 10^3. It would be a simple Dynamic programming question.


can we optimize it?


2 Answers 2


This is only a sketch of solution (there might be some off-by-ones)

Looking at a permutation of $\{1\ldots,n\}$ is equivalent at looking its inversion table $(a_1, \ldots, a_n)$ where $a_i$ is the number of elements to the left of $i$ that are greater than $i$. Basically that gives you a bijection between $S_n$ and $\prod_{1\le i\le n} \{0,\ldots,i-1\}$ and the number of inversions of the permutation is exactly the sum of all $a_i$ corresponding to it (see for example Knuth TAOCP volume 3 for a nice development).

So the problem is equivalent to computing the cardinal of:

$$\{(a_1, \ldots, a_n) \in \prod_{i=1}^{n}\{0,\ldots,i-1\}, \sum_{i=1}^n a_i=k\}$$

This way, we find a DP algorithm with $n\times k$ states:



  • When $n$ is small ($n\le k+1$):


So you can compute these in $O(k^2)$. In particular, we will need the following vector of size $k$:

$$\begin{bmatrix} \text{dp}(k+1,0)\\ \vdots\\ \text{dp}(k+1,k) \end{bmatrix}$$

  • When $n$ is large ($n>k+1$):

$$\text{dp}(n,k)=\sum_{i=0}^{k}\text{dp}(n-1,k-i)=\sum_{i=0}^k \text{dp}(n-1,i)$$

In this case $\text{dp}(n,\cdot)$ depends linearly on $\text{dp}(n-1,\cdot)$, the matrix being a $k\times k$ lower triangular matrix with nonzero coefficients being all 1-s. More formally, you can rewrite the previous equality as:

$$\begin{bmatrix} \text{dp}(n,0)\\ \vdots\\ \text{dp}(n,k) \end{bmatrix}= \begin{bmatrix} 1&0&\ldots&\ldots&0\\ 1&1&0&\ldots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 1&1&\ldots&1&0\\ 1&1&\ldots&1&1 \end{bmatrix} \begin{bmatrix} \text{dp}(n-1,0)\\ \vdots\\ \text{dp}(n-1,k) \end{bmatrix}$$

Just let me note:

$$A=\begin{bmatrix} 1&0&\ldots&\ldots&0\\ 1&1&0&\ldots&0\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ 1&1&\ldots&1&0\\ 1&1&\ldots&1&1 \end{bmatrix}$$

Now if you iterate the relation, you will get for any $i$:

$$\begin{bmatrix} \text{dp}(n+i,0)\\ \vdots\\ \text{dp}(n+i,k) \end{bmatrix}= A^i \begin{bmatrix} \text{dp}(n,0)\\ \vdots\\ \text{dp}(n,k) \end{bmatrix}$$

In particular:

$$\begin{bmatrix} \text{dp}(n,0)\\ \vdots\\ \text{dp}(n,k) \end{bmatrix}= A^{n-k-1} \begin{bmatrix} \text{dp}(k+1,0)\\ \vdots\\ \text{dp}(k+1,k) \end{bmatrix}$$

The vector on the left hand side is what we want ($\text{dp}(n,k)$), the vector on the right hand side is what we computed in the first step (when $n$ is small). So if we can compute the $k\times k$ matrix $A^{n-k-1}$, computing the answer can be done in $O(k^2)$.

Now let's see how fast we can compute $A^{n-k-1}$. I suggest three different ideas (I assume that you want the answer modulo some prime $p$):

1) Matrix exponentiation: if we can multiply $k\times k$ matrices with coefficients in $\mathbb{Z}_p$ in time $O(k^\omega)$), you can use fast exponentiation to have a $O(k^\omega \log n)$ algorithm ($\omega =2.38$ nowadays). Actually, if you are fine with a $O(k^3 \log n)$ algorithm, this should be easy to code (with the naive matrix multiplication algorithm)

2) Polynomial interpolation: if you look at the coefficients of $(A^T)^n$, it looks like:

$$(A^T)^n=\begin{bmatrix} P_1(n)&P_2(n)&\ldots&P_k(n)\\ 0&P_!(n)&\ldots&P_{k-1}(n)\\ \vdots&\ddots&\ddots&\vdots\\ 0&\ldots&0&P_1(n) \end{bmatrix}$$

where $P_1, \ldots, P_k$ are polynomials with integer coefficients and $P_i$ is of degre $i-1$. They satisfy some kind of "generalized Pascal triangle relationship":


So it's possible to compute $P_i(j)$ for $1\le i \le k$ and $0\le j\le k-1$ in $O(k^2)$. We can find coefficients of $P_i$ by doing a Newton interpolation in $O(k \log k)$, so this should give a $O(k^2 \log k)$ algorithm.

3) Find a closed form for $A^i$ with the help of some well-known combinatorial objects. As I said in the comments, the polynomials $P_1, \ldots, P_k$ are close to Faulhaber's polynomials which can be computed in linear time. This might require some heavy handmade computation. I think it should lead to a $O(k^2)$ algorithm.

  • $\begingroup$ I did not understand from "the matrix being a k×k ...". Could you please give me a little example? $\endgroup$
    – user102578
    Oct 6, 2019 at 21:40
  • $\begingroup$ @Learner007 You're right to be suspicious, because it's wrong indeed: it's a lower triangular matrix consisting only of 1 (if you write the vector $x_n=(\text{dp}(n,0),\ldots,\text{dp}(n,k))$, you have $x_n=A x_{n-1}$). So my answer is actually not very satisfying as such: you'll have a $O(k^3\log n)$ algorithm (or $O(k^{2.37}\log n)$ if that's a theoretical question), I need to sleep right now but meanwhile you can investigate what powers of this matrix $A$ are like (you'll get $k$ interesting polynomials in $n$ of degree $\le k$, not sure whether they are easy to compute) $\endgroup$
    – md5
    Oct 6, 2019 at 23:27
  • $\begingroup$ My conjecture would be that the coefficients of these polynomials are computable in $O(k^2)$, because they are very similar to the Faulhaber's polynomials (en.wikipedia.org/wiki/Faulhaber%27s_formula), which only involve Bernoulli's numbers $\endgroup$
    – md5
    Oct 6, 2019 at 23:41
  • $\begingroup$ if I want to use this method N=6 and K=3 can you explain using this example? $\endgroup$
    – user102578
    Oct 7, 2019 at 10:45
  • $\begingroup$ @Learner007: I tried to be a little bit more precise (also I added description of a $O(k^2 \log k)$ algorithm). Tell me if that's still unclear (tbh matrices in LaTeX are a bit of a pain to write) $\endgroup$
    – md5
    Oct 7, 2019 at 12:33

You could look at the answers to this question on SO: https://stackoverflow.com/questions/19372991/number-of-n-element-permutations-with-exactly-k-inversions (in particular the part about Mahonian numbers)

Additionally, it seems that dynamic programming with memoisation runs in $\mathcal{O}(N\cdot k)$ [NB: I didn't actually check this, it's in one of the answers to the linked question]: if this is for a competitive programming problem, you might consider trying it: even though in theory $N\cdot k$ can be as large as $10^{12}$, the judge system may test/query only reasonable cases (it happens).


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