# 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 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:pqduabcdqExplanation:

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:acAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

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**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