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Suffix Array

Given a string s, sort all cyclic shifts of s. Formally produce a table p such that p[j]=i if s[i:]+s[:i] has rank j among all cyclic shifts.


On input “bobocel”, here are all cyclic shifts, together with by how much they are shifted.

bobocel 0
obocelb 1
bocelbo 2
ocelbob 3
celbobo 4
elboboc 5
lboboce 6

And the same list, but lexicographically sorted. Together with the rank in the new order.

0 bobocel 0
1 bocelbo 2
2 celbobo 4
3 elboboc 5
4 lboboce 6
5 obocelb 1
6 ocelbob 3

Hence on this example we should output p=[0, 2, 4, 5, 6, 1, 3].


The naïve algorithm for this problem stores all cyclic shifts in a list and sorts it. This takes time $O(n^2 \log n)$, because the list has size $n$, and comparing two strings of length $n$ takes time $O(n)$.

The problem can be solved in time $O(n)$, under some conditions on the alphabet. But we present an $O(n \log^2 n)$ implementation, which is good enough for most programming contests.

The key operation

The algorithm relies on a simple sorting function sort_class, which not only sorts a given string or list s, but also returns additional informations. It returns two tables p and c such that

  • p[j]=i if s[i] has rank j in sorted(s).
  • c[j]=i if s[i] has rank j in sorted(set(s)).

Note that the second table groups identical elements in s, and gives a rank only to the equivalence classes. For example

index   =  0123456
input s = "bobocel"

after sorting s we have for example (because identical letters can be ordered arbitrarily within each other)

original index =  0245613
sorted s       = "bbceloo"

If we rank all distinct letters of the input string we have

rank           = 0 1 2 3 4
sorted(set(s)) = b c e l o

Hence our function returns

p = [0, 2, 4, 5, 6, 1, 3]
c = [0, 4, 0, 4, 1, 2, 3]
s =  b  o  b  o  c  e  l  # for comparison

The cyclic version

We present here the cyclic version of the problem. If we want to sort the suffixes of a given string s, then we can just sort the cyclic shifts of s + special, where special is a dummy character, smaller than all characters in s.


The idea is that for K being every integer power of 2, we want to sort the K-lengths prefixes of all cyclic shifts. For example for K=2 and s=”bobocel” we want to sort the following strings.


But we already have the order and equivalence classes of all (K/2)-length prefixes of all cyclic shifts. Now every K-lengths prefix is the concatenation of two (K/2)-length prefixes, say xy. Let i be the equivalence class of x and j be the equivalence class of y. Then the pair (i,j) represents the equivalence class of xy. We can use sort_class to translate the pairs into rank integers.

The strings of the example above correspond to the following pairs, where the table c from the previous iteration makes the correspondance.

(0, 4)
 (4, 0)
  (0, 4)
   (4, 1)
    (1, e)
     (e, 3)
      (3, 0)

In total we have a logarithmic number of outer iterations, each costs O(n log n), leading to the claimed complexity.


def sort_class(s):
    """ sorts s and returns additional information

    :param s: string or list
    :returns p, c: p[j]=i if s[i] has rank j in sorted(s) and c[i] is rank of s[i] in sorted(set(s))
    :complexity: O(n log n) or better if sort makes use of specific values in s
    S_index = [(x, i) for i, x in enumerate(s)]
    p = [i for x, i in sorted(S_index)]
    x2c = {x : i for i, x in enumerate(sorted(set(s)))}
    c = [x2c[x] for x in s]
    return p, c

def sort_cyclic_shifts(s):
    """ given a string s, sort lexicographically all cyclic shifts of s.

    The i-th cyclic shift of s is s[i:] + s[i:]
    :param s: string or list
    :returns L: such that L[j]=i if the i-th cyclic shift of s has rank j
    :complexity: O(n * log(n)^2)
    p, c = sort_class(s)
    n = len(s)
    K = 1
    while K <= n:
        L = [(c[i], c[(i + K) % n]) for i in range(n)]
        p, c = sort_class(L)
        K <<= 1 
    return p

def suffix_array(s):
    """ given a string s, sort lexicographically suffixes of s
    :param s: string
    :returns: R with R[i] is j such that s[j:] has rank i
    :complexity: O(n log^2 n)
    special = chr(0)
    assert special < min(s) 
    L = sort_cyclic_shifts(s + special)
    return L[1:]