String matching with * (that matches with any) in any of the two strings

You are given two strings A and B. Strings also contains special character * . you can replace * with any alphabetic character. Finally, you have to tell whether it is possible to make both string same or not.

Examples:

Input : A = "gee*sforgeeks"
        B = "geeksforgeeks"
Output :Yes
Input :A = "abs*"
       B = "abds"
Output :No

Explanation: How we can solve above problem, Basically we three cases, 

• Case 1: Both strings contain * at a particular position, at that time we can replace both * with any character to make the string equal at that position. 
• Case 2: If one string has character and the other has * at that position. So, we can replace * with the same character in another string. 
• Case 3: If both strings has a character at that position, then they must be same, otherwise we can’t make them equal. 

Implementation:

C++

// CPP program for string matching with *
#include <bits/stdc++.h>
using namespace std;

bool doMatch(string A, string B)
{
    for (int i = 0; i < A.length(); i++)

        // if the string don't have *
        // then character at that position
        // must be same.
        if (A[i] != '*' && B[i] != '*')
            if (A[i] != B[i])
               return false;

    return true;
}

int main()
{
    string A = "gee*sforgeeks";
    string B = "geeksforgeeks";
    cout << doMatch(A, B);
    return 0;
}

Java

// Java program for string matching with *
import java.util.*;

public class GfG {

    // Function to check if the two
    // strings can be matched or not
    public static int doMatch(String A, String B) {

        for (int i = 0; i < A.length(); i++){

            // if the string don't have *
            // then character at that position
            // must be same.
            if (A.charAt(i) != '*' && B.charAt(i) != '*'){
                if (A.charAt(i) != B.charAt(i))
                   return 0;
            }
        }

        return 1;
    }

    // Driver code
    public static void main(String []args){

        String A = "gee*sforgeeks";
        String B = "geeksforgeeks";
        System.out.println(doMatch(A, B));
    }
}

// This code is contributed by Rituraj Jain

Python3

# Python3 program for string
# matching with *

def doMatch(A, B):

    for i in range(len(A)):

        # if the string don't have *
        # then character t that position
        # must be same.
        if A[i] != '*' and B[i] != '*':
            if A[i] != B[i]:
                return False
    return True

#Driver code
if __name__=='__main__':
    A = "gee*sforgeeks"
    B = "geeksforgeeks"
    print(int(doMatch(A, B)))

# this code is contributed by
# Shashank_Sharma

C#

// C# program for string matching with
using System;

class GfG
{

    // Function to check if the two
    // strings can be matched or not
    public static int doMatch(String A, String B)
    {

        for (int i = 0; i < A.Length; i++)
        {

            // if the string don't have *
            // then character at that position
            // must be same.
            if (A[i] != '*' && B[i] != '*')
                if (A[i] != B[i])
                    return 0;
        }

        return 1;
    }

    // Driver code
    public static void Main(String []args)
    {
        String A = "gee*sforgeeks";
        String B = "geeksforgeeks";
        Console.WriteLine(doMatch(A, B));
    }
}

// This code contributed by Rajput-Ji

Javascript

<script>
// javascript program for string matching with * public class GfG {

    // Function to check if the two
    // strings can be matched or not
    function doMatch(A, B)
    {

        for (i = 0; i < A.length; i++)
        {

            // if the string don't have *
            // then character at that position
            // must be same.
            if (A.charAt(i) != '*' && B.charAt(i) != '*')
            {
                if (A.charAt(i) != B.charAt(i))
                    return 0;
            }
        }

        return 1;
    }

    // Driver code
        var A = "gee*sforgeeks";
        var B = "geeksforgeeks";
        document.write(doMatch(A, B));

// This code is contributed by aashish1995.
</script>

PHP

<?php
// PHP program for string matching with *

function doMatch($A, $B)
{
    for ($i = 0; $i < strlen($A); $i++)

        // if the string don't have *
        // then character at that position
        // must be same.
        if ($A[$i] != '*' && $B[$i] != '*')
            if ($A[$i] != $B[$i])
            return false;

    return true;
}

// Driver Code
$A = "gee*sforgeeks";
$B = "geeksforgeeks";
echo doMatch($A, $B);

// This code is contributed by Tushil.
?>

Output

1

Time Complexity: O(N)

Auxiliary Space: O(1)

Two pointer Approach:

Steps:

• If the lengths of strings A != B, return No because it’s not possible to make them the same.
• Initialize two pointers, i=0 and j= 0, pointing to the first characters of strings A and B, respectively.
• Iterate while both pointers are within the string bounds:
  1. If the characters at pointers i and j are equal or either of them is , move both pointers to the next character.
  2. If the character at pointer i is , move pointer j until the characters at pointers i and j are equal or until pointer j reaches the end of string B.
  3. If pointer j reaches the end of string B and pointer i is not pointing to , return No because it’s not possible to make them the same.
  4. If the characters at pointers i and j are not equal and pointer i is not , return No because it’s not possible to make them the same.
• If both pointers have reached the end of their respective strings, return Yes because it’s possible to make them the same.
• If either pointer i or j has not reached the end of its string, return No because it’s not possible to make them the same.

Below is the implementation of above approach:

C++

// C++ implementation of the above approach
#include <iostream>
using namespace std;

bool isPossibleToMakeSame(string A, string B)
{
    int lenA = A.length();
    int lenB = B.length();

    if (lenA != lenB)
        return false;

    int i = 0, j = 0;
    while (i < lenA && j < lenB) {
        if (A[i] == B[j] || A[i] == '*' || B[j] == '*') {
            i++;
            j++;
        }
        else if (A[i] == '*' && B[j] != '*') {
            while (B[j] != '*' && j < lenB)
                j++;
        }
        else if (A[i] != '*' && B[j] == '*') {
            while (A[i] != '*' && i < lenA)
                i++;
        }
        else {
            return false;
        }
    }

    if (i == lenA && j == lenB)
        return true;
    else
        return false;
}

// Driver Code
int main()
{
    string A = "abs*", B = "abds";

    if (isPossibleToMakeSame(A, B))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;

    return 0;
}

Java

public class GFG {

    // Function to check if it is possible to make two
    // strings equal considering '*' as a wildcard character
    public static boolean isPossibleToMakeSame(String A,
                                               String B)
    {
        int lenA = A.length();
        int lenB = B.length();

        // If the lengths of the two strings are different,
        // they cannot be made equal
        if (lenA != lenB)
            return false;

        int i = 0, j = 0;
        while (i < lenA && j < lenB) {
            // If characters at the current positions match
            // or either of them is '*', move to the next
            // positions in both strings.
            if (A.charAt(i) == B.charAt(j)
                || A.charAt(i) == '*'
                || B.charAt(j) == '*') {
                i++;
                j++;
            }
            // If character in string A is '*', skip
            // characters in string B until the next '*'.
            else if (A.charAt(i) == '*'
                     && B.charAt(j) != '*') {
                while (B.charAt(j) != '*' && j < lenB)
                    j++;
            }
            // If character in string B is '*', skip
            // characters in string A until the next '*'.
            else if (A.charAt(i) != '*'
                     && B.charAt(j) == '*') {
                while (A.charAt(i) != '*' && i < lenA)
                    i++;
            }
            // If none of the conditions above are met, the
            // strings cannot be made equal.
            else {
                return false;
            }
        }

        // If both strings are completely traversed and are
        // equal in length, they can be made equal.
        return i == lenA && j == lenB;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String A = "abs*";
        String B = "abds";

        // Check if it is possible to make the strings equal
        // considering '*' as wildcard
        if (isPossibleToMakeSame(A, B))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by shivamgupta310570

Python3

def isPossibleToMakeSame(A, B):
    lenA = len(A)
    lenB = len(B)

    # Check if the lengths of A and B are equal
    if lenA != lenB:
        return False

    i = 0
    j = 0
    while i < lenA and j < lenB:
        if A[i] == B[j] or A[i] == '*' or B[j] == '*':
            i += 1
            j += 1
        elif A[i] == '*' and B[j] != '*':
            # Skip characters in B until '*' is encountered or end of B
            while j < lenB and B[j] != '*':
                j += 1
        elif A[i] != '*' and B[j] == '*':
            # Skip characters in A until '*' is encountered or end of A
            while i < lenA and A[i] != '*':
                i += 1
        else:
            return False

    # Check if both A and B are fully traversed
    if i == lenA and j == lenB:
        return True
    else:
        return False

# Driver code
A = "abs*"
B = "abds"

if isPossibleToMakeSame(A, B):
    print("Yes")
else:
    print("No")

# THIS CODE IS CONTRIBUTED BY KANCHAN AGARWAL

C#

using System;
public class GFG
{
    // Function to check if it is possible to make two
    // strings equal considering '*' as a wildcard character
    public static bool IsPossibleToMakeSame(string A, string B)
    {
        int lenA = A.Length;
        int lenB = B.Length;

        // If the lengths of the two strings are different,
        // they cannot be made equal
        if (lenA != lenB)
            return false;

        int i = 0, j = 0;
        while (i < lenA && j < lenB)
        {
            // If characters at the current positions match
            // or either of them is '*', move to the next
            // positions in both strings.
            if (A[i] == B[j]
                || A[i] == '*'
                || B[j] == '*')
            {
                i++;
                j++;
            }
            // If character in string A is '*', skip
            // characters in string B until the next '*'.
            else if (A[i] == '*'
                     && B[j] != '*')
            {
                while (B[j] != '*' && j < lenB)
                    j++;
            }
            // If character in string B is '*', skip
            // characters in string A until the next '*'.
            else if (A[i] != '*'
                     && B[j] == '*')
            {
                while (A[i] != '*' && i < lenA)
                    i++;
            }
            // If none of the conditions above are met, the
            // strings cannot be made equal.
            else
            {
                return false;
            }
        }

        // If both strings are completely traversed and are
        // equal in length, they can be made equal.
        return i == lenA && j == lenB;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        string A = "abs*";
        string B = "abds";

        // Check if it is possible to make the strings equal
        // considering '*' as wildcard
        if (IsPossibleToMakeSame(A, B))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

Javascript

function isPossibleToMakeSame(A, B) {
    let lenA = A.length;
    let lenB = B.length;

    if (lenA !== lenB)
        return false;

    let i = 0, j = 0;
    while (i < lenA && j < lenB) {
        if (A[i] === B[j] || A[i] === '*' || B[j] === '*') {
            i++;
            j++;
        }
        else if (A[i] === '*' && B[j] !== '*') {
            while (B[j] !== '*' && j < lenB)
                j++;
        }
        else if (A[i] !== '*' && B[j] === '*') {
            while (A[i] !== '*' && i < lenA)
                i++;
        }
        else {
            return false;
        }
    }

    if (i === lenA && j === lenB)
        return true;
    else
        return false;
}

// Driver code
let A = "abs*";
let B = "abds";

if (isPossibleToMakeSame(A, B))
    console.log("Yes");
else
    console.log("No");

// THIS CODE IS CONTRIBUTED BY KANCHAN AGARWAL

Output

No

Time Complexity: O(N), where N is the length of the input strings A and B.

Auxiliary Space: O(1)