Find a string which matches all the patterns in the given array

Given an array of strings arr[] which contains patterns of characters and "*" denoting any set of characters including empty string. The task is to find a string that matches all the patterns in the array.

Note: If there is no such possible pattern, print -1.
Examples:

Input: arr[] = {“pq*du*q”, “pq*abc*q”, “p*d*q”}
Output: pqduabcdq
Explanation:
Pattern “pqduabcdq” matches all the strings:
String 1:
First “*” can be replaced by a empty string.
Second “*” can be replaced by “abcdq”.
String 2:
First “*” can be replaced by “du”
Second “*” can be replaced by “d”
String 3:
First “*” can be replaced by “q”
Second “*” can be replaced by “uabcd”

Input: arr[] = {“a*c”, “*”}
Output: ac

Approach: The idea is to find the common prefix and suffix string in all the patterns and then the middle of every pattern can be replaced by the first or last “*” in every pattern. Below is the illustration of the approach:



  • Create three empty strings that match with the prefix, suffix and middle portion of every pattern.
  • Iterate over every pattern in the array, For every pattern:
    • Find the first and last “*” in the pattern.
    • Iterate over the existing prefix and check that if it matches with the current prefix of the pattern, If any character doesn’t match with the pattern, return -1.
    • If there is some portion which is leftover in the prefix of the current pattern, then append it to the prefix of the common string.
    • Similarly, match the suffix of the pattern.
    • Finally append all the middle characters of the string except the “*” in the middle initialized string for the common string.
  • Finally, concatenate the portions of the common string that is the prefix, middle and the suffix portion of the string.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// string which matches 
// all the patterns
  
#include<bits/stdc++.h>
using namespace std;
  
// Function to find a common string 
// which matches all the pattern
string find(vector<string> S,int N)
{
    // For storing prefix till
    // first most * without conflicts
    string pref;
      
    // For storing suffix till
    // last most * without conflicts
    string suff;
      
    // For storing all middle 
    // characters between 
    // first and last *
    string mid;
          
    // Loop to iterate over every 
    // pattern of the array
    for (int i = 0; i < N; i++) {
          
        // Index of the first "*"
        int first = int(
            S[i].find_first_of('*')
            );
             
        // Index of Last "*"
        int last = int(
            S[i].find_last_of('*')
            );
          
        // Iterate over the first "*"
        for (int z = 0; z < int(pref.size()) && 
                              z < first; z++) {
            if (pref[z] != S[i][z]) {
                return "*";
            }
        }
          
        // Prefix till first most *
        // without conflicts
        for (int z = int(pref.size()); 
                       z < first; z++) {
            pref += S[i][z];
        }
          
        // Iterate till last 
        // most * from last
        for (int z = 0; z < int(suff.size()) &&
               int(S[i].size())-1-z > last; z++) {
            if (suff[z] != S[i][int(S[i].size())-1-z]) {
                return "*";
            }
        }
          
        // Make suffix till last 
        // most * without conflicts
        for (int z = int(suff.size()); 
         int(S[i].size())-1-z > last; z++) {
            suff += S[i][int(S[i].size())-1-z];
        }
          
        // Take all middle characters 
        // in between first and last most *
        for (int z = first; z <= last; z++) {
            if (S[i][z] != '*') mid += S[i][z];
        }
    }
      
    reverse(suff.begin(), suff.end());
    return pref + mid + suff;
}
  
// Driver Code
int main() {
  
    int N = 3; 
    vector<string> s(N);
      
    // Take all 
    // the strings
    s[0]="pq*du*q";
    s[1]="pq*abc*q";
    s[2]="p*d*q";
      
    // Method for finding
    // common string
    cout<<find(s,N);
      
    return 0;
}

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Output:


pqduabcdq







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