Given an input text and an array of k words, arr[], find all occurrences of all words in the input text. Let n be the length of text and m be the total number characters in all words, i.e. m = length(arr[0]) + length(arr[1]) + .. + O(n + length(arr[k1]). Here k is total numbers of input words.
Example:
Input: text = "ahishers" arr[] = {"he", "she", "hers", "his"} Output: Word his appears from 1 to 3 Word he appears from 4 to 5 Word she appears from 3 to 5 Word hers appears from 4 to 7
If we use a linear time searching algorithm like KMP, then we need to one by one search all words in text[]. This gives us total time complexity as O(n + length(word[0]) + O(n + length(word[1]) + O(n + length(word[2]) + … O(n + length(word[k1]). This time complexity can be written as O(n*k + m).
AhoCorasick Algorithm finds all words in O(n + m + z) time where z is total number of occurrences of words in text. The Aho–Corasick string matching algorithm formed the basis of the original Unix command fgrep.
 Prepocessing : Build an automaton of all words in arr[] The automaton has mainly three functions:
Go To : This function simply follows edges of Trie of all words in arr[]. It is represented as 2D array g[][] where we store next state for current state and character. Failure : This function stores all edges that are followed when current character doesn't have edge in Trie. It is represented as 1D array f[] where we store next state for current state. Output : Stores indexes of all words that end at current state. It is represented as 1D array o[] where we store indexes of all matching words as a bitmap for current state.
 Matching : Traverse the given text over built automaton to find all matching words.
Preprocessing:

We first Build a Trie (or Keyword Tree) of all words.
This part fills entries in goto g[][] and output o[].  Next we extend Trie into an automaton to support linear time matching.
This part fills entries in failure f[] and output o[].
Go to :
We build Trie. And for all characters which don’t have an edge at root, we add an edge back to root.
Failure :
For a state s, we find the longest proper suffix which is a proper prefix of some pattern. This is done using Breadth First Traversal of Trie.
Output :
For a state s, indexes of all words ending at s are stored. These indexes are stored as bitwise map (by doing bitwise OR of values). This is also computing using Breadth First Traversal with Failure.
Below is C++ implementation of AhoCorasick Algorithm
// C++ program for implementation of Aho Corasick algorithm // for string matching using namespace std; #include <bits/stdc++.h> // Max number of states in the matching machine. // Should be equal to the sum of the length of all keywords. const int MAXS = 500; // Maximum number of characters in input alphabet const int MAXC = 26; // OUTPUT FUNCTION IS IMPLEMENTED USING out[] // Bit i in this mask is one if the word with index i // appears when the machine enters this state. int out[MAXS]; // FAILURE FUNCTION IS IMPLEMENTED USING f[] int f[MAXS]; // GOTO FUNCTION (OR TRIE) IS IMPLEMENTED USING g[][] int g[MAXS][MAXC]; // Builds the string matching machine. // arr  array of words. The index of each keyword is important: // "out[state] & (1 << i)" is > 0 if we just found word[i] // in the text. // Returns the number of states that the built machine has. // States are numbered 0 up to the return value  1, inclusive. int buildMatchingMachine(string arr[], int k) { // Initialize all values in output function as 0. memset(out, 0, sizeof out); // Initialize all values in goto function as 1. memset(g, 1, sizeof g); // Initially, we just have the 0 state int states = 1; // Construct values for goto function, i.e., fill g[][] // This is same as building a Trie for arr[] for (int i = 0; i < k; ++i) { const string &word = arr[i]; int currentState = 0; // Insert all characters of current word in arr[] for (int j = 0; j < word.size(); ++j) { int ch = word[j]  'a'; // Allocate a new node (create a new state) if a // node for ch doesn't exist. if (g[currentState][ch] == 1) g[currentState][ch] = states++; currentState = g[currentState][ch]; } // Add current word in output function out[currentState] = (1 << i); } // For all characters which don't have an edge from // root (or state 0) in Trie, add a goto edge to state // 0 itself for (int ch = 0; ch < MAXC; ++ch) if (g[0][ch] == 1) g[0][ch] = 0; // Now, let's build the failure function // Initialize values in fail function memset(f, 1, sizeof f); // Failure function is computed in breadth first order // using a queue queue<int> q; // Iterate over every possible input for (int ch = 0; ch < MAXC; ++ch) { // All nodes of depth 1 have failure function value // as 0. For example, in above diagram we move to 0 // from states 1 and 3. if (g[0][ch] != 0) { f[g[0][ch]] = 0; q.push(g[0][ch]); } } // Now queue has states 1 and 3 while (q.size()) { // Remove the front state from queue int state = q.front(); q.pop(); // For the removed state, find failure function for // all those characters for which goto function is // not defined. for (int ch = 0; ch <= MAXC; ++ch) { // If goto function is defined for character 'ch' // and 'state' if (g[state][ch] != 1) { // Find failure state of removed state int failure = f[state]; // Find the deepest node labeled by proper // suffix of string from root to current // state. while (g[failure][ch] == 1) failure = f[failure]; failure = g[failure][ch]; f[g[state][ch]] = failure; // Merge output values out[g[state][ch]] = out[failure]; // Insert the next level node (of Trie) in Queue q.push(g[state][ch]); } } } return states; } // Returns the next state the machine will transition to using goto // and failure functions. // currentState  The current state of the machine. Must be between // 0 and the number of states  1, inclusive. // nextInput  The next character that enters into the machine. int findNextState(int currentState, char nextInput) { int answer = currentState; int ch = nextInput  'a'; // If goto is not defined, use failure function while (g[answer][ch] == 1) answer = f[answer]; return g[answer][ch]; } // This function finds all occurrences of all array words // in text. void searchWords(string arr[], int k, string text) { // Preprocess patterns. // Build machine with goto, failure and output functions buildMatchingMachine(arr, k); // Initialize current state int currentState = 0; // Traverse the text through the nuilt machine to find // all occurrences of words in arr[] for (int i = 0; i < text.size(); ++i) { currentState = findNextState(currentState, text[i]); // If match not found, move to next state if (out[currentState] == 0) continue; // Match found, print all matching words of arr[] // using output function. for (int j = 0; j < k; ++j) { if (out[currentState] & (1 << j)) { cout << "Word " << arr[j] << " appears from " << i  arr[j].size() + 1 << " to " << i << endl; } } } } // Driver program to test above int main() { string arr[] = {"he", "she", "hers", "his"}; string text = "ahishers"; int k = sizeof(arr)/sizeof(arr[0]); searchWords(arr, k, text); return 0; }
Output:
Word his appears from 1 to 3 Word he appears from 4 to 5 Word she appears from 3 to 5 Word hers appears from 4 to 7
Source:
http://www.cs.uku.fi/~kilpelai/BSA05/lectures/slides04.pdf
This article is contributed by Ayush Govil. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above