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

Last Updated : 15 Oct, 2020

Given an array of strings arr[] which contains patterns of characters and “*” denoting any set of characters including the 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
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 experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 the current prefix of the pattern, If any character doesn’t match with the pattern, return -1.
- If there is some portion that 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++

// 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;
}

Python3

# Python3 implementation
# to find the string which
# matches all the patterns
 
# Function to find a common
# string which matches all
# the pattern
def find(S, N):
 
    # For storing prefix
    # till first most *
    # without conflicts
    pref = ""
 
    # For storing suffix
    # till last most *
    # without conflicts
    suff = ""
 
    # For storing all middle
    # characters between
    # first and last *
    mid = ""
 
    # Loop to iterate over every
    # pattern of the array
    for i in range(N):
 
        # Index of the first "*"
        first = int(S[i].index("*"))
 
        # Index of Last "*"
        last = int(S[i].rindex("*"))
 
        # Iterate over the first "*"
        for z in range(len(pref)):
            if(z < first):
                if(pref[z] != S[i][z]):
                    return "*"
 
        # Prefix till first most *
        # without conflicts
        for z in range(len(pref),first):
            pref += S[i][z];
 
        # Iterate till last
        # most * from last
        for z in range(len(suff)):
            if(len(S[i]) - 1 - z > last):
                if(suff[z] != S[i][len(S[i]) - 1 - z]):
                    return "*"
 
        # Make suffix till last
        # most * without conflicts
        for z in range(len(suff),
                       len(S[i]) - 1 - last):
            suff += S[i][len(S[i]) - 1 - z]
 
        # Take all middle characters
        # in between first and last most *
        for z in range(first, last + 1):
            if(S[i][z] != '*'):
                mid += S[i][z]
     
    suff=suff[:: -1]
    return pref + mid + suff
 
# Driver Code
N = 3
s = ["" for i in range(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
print(find(s, N))
 
# This code is contributed by avanitrachhadiya2155

Output:

pqduabcdq

My Personal Notes

Related Articles

Related Articles

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

C++

Python3