Skip to content
Related Articles

Related Articles

Improve Article
Match a pattern and String without using regular expressions
  • Difficulty Level : Hard
  • Last Updated : 04 Jun, 2021

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

 

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 –
 

CPP




// C++ program to find out if string follows
// a given pattern or not
#include <bits/stdc++.h>
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;
}

Java




import java.util.HashMap;
import java.util.Map;
 
class Main
{
  // Function to determine if given pattern matches with a string or not
  public static boolean match(String str, int i,
                              String pat, int j,
                              Map<Character, String> map)
  {
    int n = str.length(), m = pat.length();
 
    // base condition
    if (n < m) {
      return false;
    }
 
    // if both pattern and the string reaches end
    if (i == n && j == m) {
      return true;
    }
 
    // if either string or pattern reaches end
    if (i == n || j == m) {
      return false;
    }
 
    // consider next character from the pattern
    char curr = pat.charAt(j);
 
    // if the character is seen before
    if (map.containsKey(curr))
    {
      String s = map.get(curr);
      int k = s.length();
 
      // ss stores next k characters of the given string
      String ss;
      if (i + k < str.length()) {
        ss = str.substring(i, i + k);
      } else {
        ss = str.substring(i);
      }
 
      // return false if next k characters doesn't match with s
      if (ss.compareTo(s) != 0) {
        return false;
      }
 
      // recur for remaining characters if next k characters matches
      return match(str, i + k, pat, j + 1, map);
    }
 
    // process all remaining characters in the string if current
    // character is never seen before
    for (int k = 1; k <= n - i; k++)
    {
      // insert substring formed by next k characters of the string
      // into the map
      map.put(curr, str.substring(i, i + k));
 
      // check if it leads to the solution
      if (match(str, i + k, pat, j + 1, map)) {
        return true;
      }
 
      // else backtrack - remove current character from the map
      map.remove(curr);
    }
 
    return false;
  }
 
  public static void main(String[] args)
  {
    // input string and pattern
    String str = "GeeksforGeeks";
    String pat = "GfG";
 
    // create a map to store mappings between the pattern and string
    Map<Character, String> map = new HashMap<>();
 
    // check for solution
    if (match(str, 0, pat, 0, map))
    {
      for (Map.Entry<Character, String> entry: map.entrySet()) {
        System.out.println(entry.getKey() + "->" + entry.getValue());
      }
    }
    else {
      System.out.println("Solution doesn't exist");
    }
  }
}
 
//This code is contributed by Priyadarshini Kumari

Python3




# Function to determine if given pattern matches with a string or not
def match(str, pat, dict, i=0, j=0):
 
    n = len(str)
    m = len(pat)
 
    # base condition
    if n < m:
        return False
 
    # if both pattern and the string reaches end
    if i == n and j == m:
        return True
 
    # if either string or pattern reaches end
    if i == n or j == m:
        return False
 
    # consider next character from the pattern
    curr = pat[j]
 
    # if the character is seen before
    if curr in dict:
 
        s = dict[curr]
        k = len(s)
 
        # ss stores next k characters of the given string
        if i + k < len(str):
            ss = str[i:i + k]
        else:
            ss = str[i:]
 
        # return false if next k characters doesn't match with s
        if ss != s:
            return False
 
        # recur for remaining characters if next k characters matches
        return match(str, pat, dict, i + k, j + 1)
 
    # process all remaining characters in the string if current
    # character is never seen before
    for k in range(1, n - i + 1):
 
        # insert substring formed by next k characters of the string
        # into the dictionary
        dict[curr] = str[i:i + k]
 
        # check if it leads to the solution
        if match(str, pat, dict, i + k, j + 1):
            return True
 
        # else backtrack - remove current character from the dictionary
        dict.pop(curr)
 
    return False
 
 
if __name__ == '__main__':
 
    # input string and pattern
    str = "GeeksforGeeks"
    pat = "GfG"
 
    # create a dictionary to store mappings between the pattern and string
    dict = {}
 
    # check for solution
    if match(str, pat, dict):
        print(dict)
    else:
        print("Solution doesn't exist")
 
# This code is contributed by Priyadarshini Kumari

Output: 
 

f->for
G->Geeks

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :