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]) + … + length(arr[k-1]). Here** k** is total numbers of input words.

Example:

Input: text = "ahishers" arr[] = {"he", "she", "hers", "his"} Output: Wordhisappears from 1 to 3 Wordheappears from 4 to 5 Wordsheappears from 3 to 5 Wordhersappears 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[k-1]). This time complexity can be written as * O(n*k + m)*.

**Aho-Corasick Algorithm**finds all words in

*time where*

**O(n + m + z)****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 arrayg[][]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 arrayf[]where we store next state for current state. Output : Stores indexes of all words that end at current state. It is represented as 1D arrayo[]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 Aho-Corasick Algorithm

## C

`// 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;` `}` |

*chevron_right*

*filter_none*

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

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