TASTR - Editorial








Suffix Array, Longest Common Prefix Array


Given two strings a and b, let A be the set of all substrings of a, and B be the set of all substrings of b.

Find the number of unique strings in A plus the number of unique strings in B, that are not common to both A and B.


Efficient solutions to a large set of problems on strings are approachable by use of Suffix Arrays (or Trees). If you have not yet added Suffix Sorting (for construction of Suffix Arrays) to your skill-set, then this is the perfect problem to do so.

Suffix Sorting can be done using

  • Manber Myers algorithm in O(L log L) time
  • Karkkainan Sander’s algorithm in O(L) time

Longest Common Prefixes of adjacent items in the list of sorted suffixes is a classical problem in strings. Most texts discuss construction of Suffix Arrays immediately followed by calculation of LCP arrays.

The LCP Array can be found by augmenting the Suffix Sorting algorithms above. The time complexity for LCP Array calculation will be exactly equal to the time complexity of the Suffix Sorting. You can find it as an entirely different step as well, following the Suffix Sorting.

Both O(L) and O(L log L) approaches will fit within the time limit for this problem. Hence, choose as you please :slight_smile:

Let us assume that we can find the number of unique sub-strings of a string by use of the LCP array for the suffixes of that string. (We will see the algorithm to do so in the EXPLANATION section)

U(A) = number of unique strings in A

We are given two strings a and b and their set of substrings A and B respectively. We wish to find

U(A) + U(B) - U(A intersection B)

Which is equal to

U(A) + U(B) - (U(A union B) - U(A) - U(B))


2*U(A) + 2*U(B) - U(A union B)


Let us see how to find the number of unique substrings of a string by using the LCP array.

  • If we consider all the prefixes of all the suffixes, we would be considering the entire set of sub-strings.
  • The LCA array helps us determine how many prefixes to ignore for each suffix
  • We of course do not ignore any prefix of the first suffix. These are all valid and unique substrings.
Let s be given string
    indexes in s are 1-based
Let S be the suffix array
    S stores the list of 1-based indexes
    that represent the start position of the suffix of s
Let L be the LCP array
    L(i) is the longest common prefix between
    the suffixes starting from S(i) and S(i-1)
    Thus, L is only defined from 2 to |s|

uniq_sub_strings = |s| - S[1] + 1
// thus we count all prefixes of the first suffix

for i = 2 to N
    uniq_sub_strings += |s| - S* + 1 - L*

Let us try this with an example to see why this is correct.

s = abaabba
S = [
    7,    // a
    3,    // aabba
    1,    // abaabba
    4,    // abba
    6,    // ba
    2,    // baabba
    5     // bba
L = [
    0,    // not defined for L[1]
    1,    // a is the common prefix
    1,    // a
    2,    // ab
    0,    // nothing
    2,    // ba
    1     // b

uniq_sub_strings =
    1 + 4 + 6 + 2 + 2 + 4 + 2 = 21

Thus, there are 21 substrings (out of 28) that are unique. You can work this out by deducing that the sub-strings that are not unique (and hence weren’t counted are)

    a,  // prefix of aabba
        // since a was counted as prefix of "a"
    a,  // prefix of abaabba
    a,  // prefix of abba
    ab, // prefix of abba
        // since ab was counted as prefix of "abaabba"
    b,  // prefix of baabba
        // since b was counted as prefix of "ba"
    ba, // prefix of baabba
        // since ba was counted as prefix of "ba"
    b   // prefix of bba

Now, you can find U(A) and U(B) by using the above algorithm. The only part that remains is to calculate U(A union B).

This can be done by considering a string

c = a + "$" + b

We can find all the unique substrings of c and reduce from the result, all the strings that contain the character '</b>. We know that all the different strings that contain <b>' are unique anyway, since ‘$’ is not part of the alphabet.

There are (|a|+1) * (|b|+1) substrings of c that contain ‘$’.

Thus, we can find U(A union B), the number of unique substrings that exists in either A, or B.

We can find the number of unique sub-strings of either a, or b, but not both, in time O(F(|a|) + F(|b|) + F(|a+b|), where F(n) is the complexity of calculating the Suffix Array or the LCA Array (which ever is more). In the best implementation, F(n) = O(n). But the problem may be solved even with F(n) = O(n log n).


Can be found here.


Can be found here.


My first solution constructed the suffix array (in O(Nlog(N)) time) for each string independently (A, B and the union of A and B) and I got Time limit exceeded. I had to change my algorithm so that it computed everything only from the suffix array (and the LCPs) of the union of A and B. I thought that was intended, but now I see that I probably have some inefficiencies in my suffix array routines, which I will have to identify.


does the solution that uses hashes exist ?


Where can i read more about suffix array and LCP array? (algorithm and application)


Amazing tutorial…gr8t work!!!



My solution works on all test cases i found

but still gives runtime error

please help


@anton_lunyov sir may u plzz provide me the test cases where this solution is failed http://www.codechef.com/viewsolution/1967060
thanx in advance…:slight_smile:


I can’t seem to visualise this part

U(A) + U(B) - U(A intersection B)
Which is equal to
U(A) + U(B) - (U(A union B) - U(A) - U(B))

Isn’t U(A intersection B) supposed to be = to U(A) + U(B) - U(A union B)


Can someone help me with the stdin and stdout for node.js. My code is as follows



process.stdin.on(‘data’, function (chunk) {
var lines = chunk.toString().split(’

var stringDiff = function(a,b){
	var aSet = getAllSubstring(a);
	var bSet = getAllSubstring(b);
	var nSet = difference(aSet, bSet);
	nSet = nSet.concat(difference(bSet, aSet));
	return nSet.length;

var difference = function(a,b){
	var nSet = [];
	for(var i in a){
		var found = false;
		for(x in b){
			if(a* === b[x]){
				found = true;
		if(!found){ nSet.push(a*)};
	return nSet;

var getAllSubstring = function(s){
	var aSet = [];
	var i = 1;
	var j = 0;
		var next = s.substring(j,i);
			i = j+1;
	return aSet;

var diff = stringDiff(lines[0], lines[1]);



@anton sir would you please tell me where is my code failing here’s my code :frowning:


@anton_lunyov : sir please help me every time i get a run time error(SIGSEGV)…



could you just tell me weather my code’s output is correct or not, irrespective of time exceed error…


why doesnt work?

module Main where
import Data.List
import System.Exit
al1 xs = nub . concat . map (drop 1 . inits) . tails $ xs
cmp xs xz = [ x | x <- ((al1 xs)++(al1 xz)), notElem x (intersect (al1 xs) (al1 xz) )] 
main = do
    line <- getLine
    line2 <- getLine
    putStr $ show $ length $ cmp line line2
    exitWith ExitSuccess


This is my code in python. it works fine on my system but dont know why it is not working in codechef



getting run time error NZEC. first submission in codechef. so dont know much abt online compilers. plz help me out sir http://www.codechef.com/viewsolution/5666193


Wow… that was a quite a sleek approach… Nice one


This problem really showed our “level” :slight_smile:


I did the same thing as the tester . My solutions (during the contest[http://www.codechef.com/viewsolution/1964089]) and after the contest with some optimizations (http://www.codechef.com/viewsolution/1964221) both TLE … Is this too strict for Java ?
EDIT: I think he has a better Suffix array function.


My O(n*(logn)*(logn)) passes !


Yes @acmonster construct LCP array using hashes in O(N log^2 N) time. We can compare two suffixes in O(log N) time using hashes. So we can sort them in O(N log^2 N) time. And then calculate LCP array in O(N log N) time.