# Match a pattern and String without using regular expressions

Given a string, find out if string follows a given pattern or not without using any regular expressions.

Examples:

```Input:
string - GraphTreesGraph
pattern - aba
Output:
a->Graph
b->Trees

Input:
string - GraphGraphGraph
pattern - aaa
Output:
a->Graph

Input:
string - GeeksforGeeks
pattern - GfG
Output:
G->Geeks
f->for

Input:
string - GeeksforGeeks
pattern - GG
Output:
No solution exists
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can solve this problem with the help of Backtracking. For each character in the pattern, if the character is not seen before, we consider all possible sub-strings and recurse to see if it leads to the solution or not. We maintain a map that stores sub-string mapped to a pattern character. If pattern character is seen before, we use the same sub-string present in the map. If we found a solution, for each distinct character in the pattern, we print string mapped to it using our map.

Below is C++ implementation of above idea –

 `// C++ program to find out if string follows ` `// a given pattern or not ` `#include ` `using` `namespace` `std; ` ` `  `/*  Function to find out if string follows a given ` `    ``pattern or not ` ` `  `    ``str - given string ` `    ``n - length of given string ` `    ``i - current index in input string ` `    ``pat - given pattern ` `    ``m - length of given pattern ` `    ``j - current index in given pattern ` `    ``map - stores mapping between pattern and string */` `bool` `patternMatchUtil(string str, ``int` `n, ``int` `i, ` `                    ``string pat, ``int` `m, ``int` `j, ` `                    ``unordered_map<``char``, string>& map) ` `{ ` `    ``// If both string and pattern reach their end ` `    ``if` `(i == n && j == m) ` `        ``return` `true``; ` ` `  `    ``// If either string or pattern reach their end ` `    ``if` `(i == n || j == m) ` `        ``return` `false``; ` ` `  `    ``// read next character from the pattern ` `    ``char` `ch = pat[j]; ` ` `  `    ``// if character is seen before ` `    ``if` `(map.find(ch)!= map.end()) ` `    ``{ ` `        ``string s = map[ch]; ` `        ``int` `len = s.size(); ` ` `  `        ``// consider next len characters of str ` `        ``string subStr = str.substr(i, len); ` ` `  `        ``// if next len characters of string str ` `        ``// don't match with s, return false ` `        ``if` `(subStr.compare(s)) ` `            ``return` `false``; ` ` `  `        ``// if it matches, recurse for remaining characters ` `        ``return` `patternMatchUtil(str, n, i + len, pat, m, ` `                                            ``j + 1, map); ` `    ``} ` ` `  `    ``// If character is seen for first time, try out all ` `    ``// remaining characters in the string ` `    ``for` `(``int` `len = 1; len <= n - i; len++) ` `    ``{ ` `        ``// consider substring that starts at position i ` `        ``// and spans len characters and add it to map ` `        ``map[ch] = str.substr(i, len); ` ` `  `        ``// see if it leads to the solution ` `        ``if` `(patternMatchUtil(str, n, i + len, pat, m, ` `                                          ``j + 1, map)) ` `            ``return` `true``; ` ` `  `        ``// if not, remove ch from the map ` `        ``map.erase(ch); ` `    ``} ` ` `  `    ``return` `false``; ` `} ` ` `  `// A wrapper over patternMatchUtil()function ` `bool` `patternMatch(string str, string pat, ``int` `n, ``int` `m) ` `{ ` `    ``if` `(n < m) ` `    ``return` `false``; ` ` `  `    ``// create an empty hashmap ` `    ``unordered_map<``char``, string> map; ` ` `  `    ``// store result in a boolean variable res ` `    ``bool` `res = patternMatchUtil(str, n, 0, pat, m, 0, map); ` ` `  `    ``// if solution exists, print the mappings ` `    ``for` `(``auto` `it = map.begin(); res && it != map.end(); it++) ` `        ``cout << it->first << ``"->"` `<< it->second << endl; ` ` `  `    ``// return result ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"GeeksforGeeks"``, pat = ``"GfG"``; ` ` `  `    ``int` `n = str.size(), m = pat.size(); ` ` `  `    ``if` `(!patternMatch(str, pat, n, m)) ` `        ``cout << ``"No Solution exists"``; ` ` `  `    ``return` `0; ` `} `

Output:

```f->for
G->Geeks
```

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

## tags:

Backtracking Backtracking

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