Aho-Corasick Algorithm

Data StructureAlgorithmsPattern Searching Algorithms

This algorithm is helpful to find all occurrences of all given set of keywords. It is a kind of Dictionary-matching algorithm. It uses a tree structure using all keywords. After making the tree, it tries to convert the tree as an automaton to make the searching in linear time. There are three different phases of Aho-Corasick Algorithm. 

These are Go-to, Failure, and Output. In the go-to stage, it makes the tree using all the keywords. In the next phase or in the Failure Phase, it tries to find the backward transition to get a proper suffix of some keywords. In the output stage, for every state ‘s’ of the automaton, it finds all words which are ending at the state ‘s’.

The time complexity of this algorithm is: O(N + L + Z), where N is the length of the text, L is the length of keywords and the Z is a number of matches.

Input and Output

A set of patterns: {their, there, answer, any, bye}
The main string: “isthereanyanswerokgoodbye”
Word there location: 2
Word any location: 7
Word answer location: 10
Word bye location: 22


buildTree(patternList, size)

Input − The list of all patterns, and the size of the list

Output − Generate the transition map to find the patterns

   set all elements of output array to 0
   set all elements of fail array to -1
   set all elements of goto matrix to -1
   state := 1       //at first there is only one state.

   for all patterns ‘i’ in the patternList, do
      word := patternList[i]
      present := 0
      for all character ‘ch’ of word, do
         if goto[present, ch] = -1 then
            goto[present, ch] := state
            state := state + 1
         present:= goto[present, ch]
      output[present] := output[present] OR (shift left 1 for i times)

   for all type of characters ch, do
      if goto[0, ch] ≠ 0 then
         fail[goto[0,ch]] := 0
         insert goto[0, ch] into a Queue q.

   while q is not empty, do
      newState := first element of q
      delete from q.
      for all possible character ch, do
         if goto[newState, ch] ≠ -1 then
            failure := fail[newState]
            while goto[failure, ch] = -1, do
               failure := goto[failure, ch]

            fail[goto[newState, ch]] = failure
            output[goto[newState, ch]] :=output[goto[newState,ch]] OR output[failure]
            insert goto[newState, ch] into q.
   return state

getNextState(presentState, nextChar)

Input − present state and the next character to determine next state

Output: the next state

   answer := presentState
   ch := nextChar

   while goto[answer, ch] = -41, do
      answer := fail[answer]
   return goto[answer, ch]

patternSearch(patternList, size, text)

Input − List of patterns, size of the list and the main text

Output − The indexes of the text where patterns are found

   call buildTree(patternList, size)
   presentState := 0

   for all indexes of the text, do
      if output[presentState] = 0
         ignore next part and go for next iteration
      for all patterns in the patternList, do
         if the pattern found using output array, then
            print the location where pattern is present


#include <iostream>
#include <queue>
#define MAXS 500    //sum of the length of all patterns
#define MAXC 26     //as 26 letters in alphabet
using namespace std;

int output[MAXS];
int fail[MAXS];
int gotoMat[MAXS][MAXC];

int buildTree(string array[], int size) {
   for(int i = 0; i<MAXS; i++)
      output[i] = 0;    //all element of output is 0

   for(int i = 0; i<MAXS; i++)
      fail[i] = -1;    //all element of failure array is -1

   for(int i = 0; i<MAXS; i++)
      for(int j = 0; j<MAXC; j++)
         gotoMat[i][j] = -1;    //all element of goto matrix is -1

   int state = 1;    //there is only one state

   for (int i = 0; i < size; i++) {    //make trie structure for all pattern in array
      //const string &word = array[i];
      string word = array[i];
      int presentState = 0;

      for (int j = 0; j < word.size(); ++j) {    //all pattern is added
         int ch = word[j] - 'a';
         if (gotoMat[presentState][ch] == -1)    //ic ch is not present make new node
            gotoMat[presentState][ch] = state++;    //increase state
            presentState = gotoMat[presentState][ch];
      output[presentState] |= (1 << i); //current word added in the output

   for (int ch = 0; ch < MAXC; ++ch)   //if ch is not directly connected to root
      if (gotoMat[0][ch] == -1)
         gotoMat[0][ch] = 0;

         queue<int> q;

   for (int ch = 0; ch < MAXC; ++ch) {    //node goes to previous state when fails
      if (gotoMat[0][ch] != 0) {
         fail[gotoMat[0][ch]] = 0;

   while (q.size()) {
      int state = q.front();    //remove front node

      for (int ch = 0; ch <= MAXC; ++ch) {
         if (gotoMat[state][ch] != -1) {    //if goto state is present
            int failure = fail[state];
            while (gotoMat[failure][ch] == -1)    //find deepest node with proper suffix
               failure = fail[failure];
            failure = gotoMat[failure][ch];
            fail[gotoMat[state][ch]] = failure;
            output[gotoMat[state][ch]] |= output[failure];   // Merge output values
            q.push(gotoMat[state][ch]);    //add next level node to the queue
   return state;

int getNextState(int presentState, char nextChar) {
   int answer = presentState;
   int ch = nextChar - 'a'; //subtract ascii of 'a'

   while (gotoMat[answer][ch] == -1) //if go to is not found, use fail function
      answer = fail[answer];
   return gotoMat[answer][ch];

void patternSearch(string arr[], int size, string text) {
   buildTree(arr, size);    //make the trie structure
   int presentState = 0;    //make current state as 0

   for (int i = 0; i < text.size(); i++) {    //find all occurances of pattern
      presentState = getNextState(presentState, text[i]);
      if (output[presentState] == 0)    //if no match continue;
      for (int j = 0; j < size; ++j) {   //matching found and print words
         if (output[presentState] & (1 << j)) {
            cout << "Word " << arr[j] << " location: " << i - arr[j].size() + 1 << endl;

int main() {
   string arr[] = {"their", "there", "answer", "any", "bye"};
   string text = "isthereanyanswerokgoodbye";
   int k = sizeof(arr)/sizeof(arr[0]);
   patternSearch(arr, k, text);
   return 0;


Word there location: 2
Word any location: 7
Word answer location: 10
Word bye location: 22
Published on 09-Jul-2018 11:49:46