Given a permutation on {0,1,…,n-1} find its rank for the lexicographical order. Given a rank find the corresponding permutation.

## Example

Say n=4, then the 4!=24 permutations are lexicographically ordered as follows.

rank permutation
0 0,1,2,3
1 0,1,3,2
2 0,2,1,3
3 0,2,3,1
4 0,3,1,2
5 0,3,2,1
….
23 3,2,1,0

## An observation

Suppose we are given a permutation $(p_{n-1},\ldots,p_1,p_0)$ and which to compute its rank.

How many permutations start with a 0? Well the remainder is a permutation over {1,..,n-1}, and all these permutations are valid completions of the initial 0. Hence there are $(n-1)!$ permutations that start with a 0. Similarly there are $(n-1)!$ permutations that start with a 1, and in the lexicographical order they come after those that start with a 0. This means that a first contribution to the rank is $p_{n-1} \cdot (n-1)!$.

The situation for the next value $p_{n-2}$ is slightly different. Again there are $(n-2)!$ permutation that start with fixed values $p_{n-1},p_{n-2}$. However there are only $n-1$ possibilities for $p_{n-2}$, namely all n values excluding $p_{n-1}$. This means that the second contribution to the rank is $r_{n-2} \cdot (n-2)!$, where $r_{n-1}$ is the rank of $p_{n-2}$ among the set $\{0,1,\ldots,n-1\} \setminus \{p_{n-1}\}$. The key is then to translate a vector $p$ defining a permutation into a vector of ranks $r$ such that $r_i$ is the rank of $p_i$ among the values $\{p_{i},\ldots,p_1,p_0\}$. For example if p=(2,0,3,1), then r=(2,0,1,0). The rank of p is then

$r_{n-1} \cdot (n-1)! + \cdots + r_2 \cdot 2! + r_1 \cdot 1! \cdot + r_0.$

Note that $r_0$ can safely be omitted in the expression since it is always zero.

The trick is then to compute the ranks $r_i$ from the values $p_i$.

## An $O(n^2)$ algorithm

For this purpose we parse the permutation p, keeping track of the values which have not yet be seen so far and search among them to determine the rank. We maintain a list digits contains all values which have not yet seen in the given permutation. Note that the last element of the permutation can be ignored, it does not provide any information for the rank.

For the inverse, decompose r into

$r_{n-1} (n-1)! + \cdots + r_2 \cdot 2! + r_1 \cdot 1! + r_0$

with $0\leq r_i \leq i$ and map each $r_i$ to the value of rank i among the values {0,…n-1} which have not seen so far.

## An $O(n \log n)$ algorithm

In order to compute the rank of a permutation we use a table called rank which maps a value $p_i$ to a rank $r_i$ as explained above. Initially the table has the identity ranks, i.e. rank[x]=x. Then after each processed value $p_i$ we need to decrement all ranks in rank between the indices $p_i$ and $n-1$. We can use a segment tree for this purpose. Then the decrement and access operations to the rank table can be done in logarithmic time.

## Variant

Let s be some string of length n. We can ask the same question about permutations of s. The difficulty is that there can be repetitions of a same letter in s. Do you know how compute the rank-permutation bijection for this variant?