ABAB
|||
BABA
||
ABBA
Given two strings, of length and of length , find the longest strings which are a substrings of both and .
A generalisation is the k-common substring problem. Given the set of strings , where and Σ. Find for each 2 ≤ ≤ , the longest strings which occur as substrings of at least strings.
You can find the lengths and starting positions of the longest common substrings of and in with the help of a generalised suffix tree. Finding them by dynamic programming costs . The solutions to the generalised problem take and ·...· time.
You can find the longest common substrings of a set of strings by building a generalised suffix tree for the strings, and then finding the deepest internal nodes which has leaf nodes from all the strings in the subtree below it. In the figure on the right you see the suffix tree for the strings "ABAB", "BABA" and "ABBA", padded with unique string terminators, to become "ABAB$0", "BABA$1" and "ABBA$2". The nodes representing "A", "B", "AB" and "BA" all have descendant leaves from all of the strings, numbered 0, 1 and 2.
Building the suffix tree takes time (if the size of the alphabet is constant). If you traverse the tree bottom up, and maintain a bit vector telling which strings are seen below each node, you can solve the k-common substring problem in time. If you prepare your suffix tree for constant time lowest common ancestor retrieval, you can solve it in time.
For the example strings "ABAB" and "BABA":
| A | B | A | B | ||
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | |
| B | 0 | 0 | 1 | 0 | 1 |
| A | 0 | 1 | 0 | 2 | 0 |
| B | 0 | 0 | 2 | 0 | 3 |
| A | 0 | 1 | 0 | 3 | 0 |
The maximal of these longest common suffixes of possible prefixes must be the longest common substrings of S and T. These are shown on diagonals, in red, in the table.
This can be extended to more than two strings by adding more dimensions to the table.
The following pseudocode finds the set of longest common substrings between two strings with dynamic programming:
function LCSubstr(S[1..m], T[1..n])
L := array(0..m, 0..n)
z := 0ret := {}
for i := 1..m
for j := 1..n
if S[i] = T[j]
if i = 1 or j = 1
L[i,j] := 1
else
L[i,j] := L[i-1,j-1] + 1
if L[i,j] > z
z := L[i,j]ret := {}
if L[i,j] = z
ret := ret ∪ {S[i-z+1..i]}
return retThis algorithm runs in time. The variable z is used to hold the length of the longest common substring found so far. The set ret is used to hold the set of strings which are of length z
The following tricks can be used to reduce the memory usage of an implementation: