Dynamic Programming | Wildcard Pattern Matching | Linear Time and Constant Space - GeeksforGeeks

28 Feb, 2022
Given a text and a wildcard pattern, find if wildcard pattern is matched with text. The matching should cover the entire text (not partial text).The wildcard pattern can include the characters ‘?’ and ‘*’:
‘?’ – matches any single character
‘*’ – Matches any sequence of characters (including the empty sequence)
Pre-requisite: Dynamic Programming | Wildcard Pattern MatchingExamples:
Text = "baaabab",
Pattern = “*****ba*****ab", output : true
Pattern = "baaa?ab", output : true
Pattern = "ba*a?", output : true
Pattern = "a*ab", output : false 
Each occurrence of ‘?’ character in wildcard pattern can be replaced with any other character and each occurrence of ‘*’ with a sequence of characters such that the wildcard pattern becomes identical to the input string after replacement.
We have discussed a solution here which has O(m x n) time and O(m x n) space complexity.For applying the optimization, we will at the first note the BASE CASE which says, if the length of the pattern is zero then answer will be true only if the length of the text with which we have to match the pattern is also zero.Algorithm:  
Let i be the marker to point at the current character of the text. Let j be the marker to point at the current character of the pattern. Let index_txt be the marker to point at the character of text on which we encounter ‘*’ in the pattern. Let index_pat be the marker to point at the position of ‘*’ in the pattern.At any instant, if we observe that txt[i] == pat[j], then we increment both i and j as no operation needs to be performed in this case.If we encounter pat[j] == ‘?’, then it resembles the case mentioned in step – (2) as ‘?’ has the property to match with any single character.If we encounter pat[j] == ‘*’, then we update the value of index_txt and index_pat as ‘*’ has the property to match any sequence of characters (including the empty sequence) and we will increment the value of j to compare next character of pattern with the current character of the text. (As character represented by i has not been answered yet).Now if txt[i] == pat[j], and we have encountered a ‘*’ before, then it means that ‘*’ included the empty sequence, else if txt[i] != pat[j], a character needs to be provided by ‘*’ so that current character matching takes place, then i needs to be incremented as it is answered now but the character represented by j still needs to be answered, therefore, j = index_pat + 1, i = index_txt + 1 (as ‘*’ can capture other characters as well), index_txt++ (as current character in text is matched).If step – (5) is not valid, that means txt[i] != pat[j], also we have not encountered a ‘*’ that means it is not possible for the pattern to match the string. (return false).Check whether j reached its final value or not, then return the final answer.
Let i be the marker to point at the current character of the text. Let j be the marker to point at the current character of the pattern. Let index_txt be the marker to point at the character of text on which we encounter ‘*’ in the pattern. Let index_pat be the marker to point at the position of ‘*’ in the pattern.
At any instant, if we observe that txt[i] == pat[j], then we increment both i and j as no operation needs to be performed in this case.
If we encounter pat[j] == ‘?’, then it resembles the case mentioned in step – (2) as ‘?’ has the property to match with any single character.
If we encounter pat[j] == ‘*’, then we update the value of index_txt and index_pat as ‘*’ has the property to match any sequence of characters (including the empty sequence) and we will increment the value of j to compare next character of pattern with the current character of the text. (As character represented by i has not been answered yet).
Now if txt[i] == pat[j], and we have encountered a ‘*’ before, then it means that ‘*’ included the empty sequence, else if txt[i] != pat[j], a character needs to be provided by ‘*’ so that current character matching takes place, then i needs to be incremented as it is answered now but the character represented by j still needs to be answered, therefore, j = index_pat + 1, i = index_txt + 1 (as ‘*’ can capture other characters as well), index_txt++ (as current character in text is matched).
If step – (5) is not valid, that means txt[i] != pat[j], also we have not encountered a ‘*’ that means it is not possible for the pattern to match the string. (return false).
Check whether j reached its final value or not, then return the final answer.
Let us see the above algorithm in action, then we will move to the coding section:text = “baaabab” pattern = “*****ba*****ab”NOW APPLYING THE ALGORITHM Step – (1) : i = 0 (i –> ‘b’) j = 0 (j –> ‘*’) index_txt = -1 index_pat = -1NOTE: LOOP WILL RUN TILL i REACHES ITS FINAL VALUE OR THE ANSWER BECOMES FALSE MIDWAY. FIRST COMPARISON :- As we see here that pat[j] == ‘*’, therefore directly jumping on to step – (4). Step – (4) : index_txt = i (index_txt –> ‘b’) index_pat = j (index_pat –> ‘*’) j++ (j –> ‘*’)After four more comparisons : i = 0 (i –> ‘b’) j = 5 (j –> ‘b’) index_txt = 0 (index_txt –> ‘b’) index_pat = 4 (index_pat –> ‘*’)SIXTH COMPARISON :- As we see here that txt[i] == pat[j], but we already encountered ‘*’ therefore using step – (5). Step – (5) : i = 1 (i –> ‘a’) j = 6 (j –> ‘a’) index_txt = 0 (index_txt –> ‘b’) index_pat = 4 (index_pat –> ‘*’)SEVENTH COMPARISON :- Step – (5) : i = 2 (i –> ‘a’) j = 7 (j –> ‘*’) index_txt = 0 (index_txt –> ‘b’) index_pat = 4 (index_pat –> ‘*’)EIGHT COMPARISON :- Step – (4) : i = 2 (i –> ‘a’) j = 8 (j –> ‘*’) index_txt = 2 (index_txt –> ‘a’) index_pat = 7 (index_pat –> ‘*’)After four more comparisons : i = 2 (i –> ‘a’) j = 12 (j –> ‘a’) index_txt = 2 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)THIRTEENTH COMPARISON :- Step – (5) : i = 3 (i –> ‘a’) j = 13 (j –> ‘b’) index_txt = 2 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)FOURTEENTH COMPARISON :- Step – (5) : i = 3 (i –> ‘a’) j = 12 (j –> ‘a’) index_txt = 3 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)FIFTEENTH COMPARISON :- Step – (5) : i = 4 (i –> ‘b’) j = 13 (j –> ‘b’) index_txt = 3 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)SIXTEENTH COMPARISON :- Step – (5) : i = 5 (i –> ‘a’) j = 14 (j –> end) index_txt = 3 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)SEVENTEENTH COMPARISON :- Step – (5) : i = 4 (i –> ‘b’) j = 12 (j –> ‘a’) index_txt = 4 (index_txt –> ‘b’) index_pat = 11 (index_pat –> ‘*’)EIGHTEENTH COMPARISON :- Step – (5) : i = 5 (i –> ‘a’) j = 12 (j –> ‘a’) index_txt = 5 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)NINETEENTH COMPARISON :- Step – (5) : i = 6 (i –> ‘b’) j = 13 (j –> ‘b’) index_txt = 5 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)TWENTIETH COMPARISON :- Step – (5) : i = 7 (i –> end) j = 14 (j –> end) index_txt = 5 (index_txt –> ‘a’) index_pat = 11 (index_pat –> ‘*’)NOTE : NOW WE WILL COME OUT OF LOOP TO RUN STEP – 7. Step – (7) : j is already present at its end position, therefore answer is true.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program to implement wildcard// pattern matching algorithm#include <bits/stdc++.h>using namespace std; // Function that matches input text// with given wildcard patternbool strmatch(char txt[], char pat[],              int n, int m){         // empty pattern can only    // match with empty string.    // Base Case :    if (m == 0)        return (n == 0);     // step-1 :    // initialize markers :    int i = 0, j = 0, index_txt = -1,                       index_pat = -1;     while (i < n)    {         // For step - (2, 5)        if (j < m && txt[i] == pat[j])        {            i++;            j++;        }         // For step - (3)        else if (j < m && pat[j] == '?')        {            i++;            j++;        }         // For step - (4)        else if (j < m && pat[j] == '*')        {            index_txt = i;            index_pat = j;            j++;        }         // For step - (5)        else if (index_pat != -1)        {            j = index_pat + 1;            i = index_txt + 1;            index_txt++;        }         // For step - (6)        else        {            return false;        }    }     // For step - (7)    while (j < m && pat[j] == '*')    {        j++;    }     // Final Check    if (j == m)    {        return true;    }     return false;} // Driver codeint main(){        char str[] = "baaabab";    char pattern[] = "*****ba*****ab";    // char pattern[] = "ba*****ab";    // char pattern[] = "ba*ab";    // char pattern[] = "a*ab";     if (strmatch(str, pattern,                 strlen(str), strlen(pattern)))        cout << "Yes" << endl;    else        cout << "No" << endl;     char pattern2[] = "a*****ab";    if (strmatch(str, pattern2,                 strlen(str), strlen(pattern2)))        cout << "Yes" << endl;    else        cout << "No" << endl;     return 0;}
// Java program to implement wildcard// pattern matching algorithmclass GFG {     // Function that matches input text    // with given wildcard pattern    static boolean strmatch(char txt[], char pat[],                            int n, int m)    {        // empty pattern can only        // match with empty string.        // Base Case :        if (m == 0)            return (n == 0);         // step-1 :        // initialize markers :        int i = 0, j = 0, index_txt = -1,            index_pat = -1;         while (i < n)        {             // For step - (2, 5)            if (j < m && txt[i] == pat[j])            {                i++;                j++;            }             // For step - (3)            else if (j < m && pat[j] == '?')            {                i++;                j++;            }             // For step - (4)            else if (j < m && pat[j] == '*')            {                index_txt = i;                index_pat = j;                j++;            }             // For step - (5)            else if (index_pat != -1)            {                j = index_pat + 1;                i = index_txt + 1;                index_txt++;            }             // For step - (6)            else            {                return false;            }        }         // For step - (7)        while (j < m && pat[j] == '*')        {            j++;        }         // Final Check        if (j == m)        {            return true;        }         return false;    }     // Driver code    public static void main(String[] args)    {                 char str[] = "baaabab".toCharArray();        char pattern[] = "*****ba*****ab".toCharArray();        // char pattern[] = "ba*****ab";        // char pattern[] = "ba*ab";        // char pattern[] = "a*ab";         if (strmatch(str, pattern, str.length,                     pattern.length))            System.out.println("Yes");        else            System.out.println("No");         char pattern2[] = "a*****ab".toCharArray();        if (strmatch(str, pattern2, str.length,                     pattern2.length))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed by Rajput-Ji
# Python3 program to implement# wildcard pattern matching# algorithm # Function that matches input# txt with given wildcard patterndef stringmatch(txt, pat, n, m):         # empty pattern can only    # match with empty sting    # Base case    if (m == 0):        return (n == 0)             # step 1    # initialize markers :    i = 0    j = 0    index_txt = -1    index_pat = -1    while(i < n - 2):                 # For step - (2, 5)        if (j < m and txt[i] == pat[j]):            i += 1            j += 1                     # For step - (3)        elif(j < m and pat[j] == '?'):            i += 1            j += 1                     # For step - (4)        elif(j < m and pat[j] == '*'):            index_txt = i            index_pat = j            j += 1                     # For step - (5)        elif(index_pat != -1):            j = index_pat + 1            i = index_txt + 1            index_txt += 1                     # For step - (6)        else:            return False    # For step - (7)    while (j < m and pat[j] == '*'):        j += 1     # Final Check    if(j == m):        return True     return False # Driver codestrr = "baaabab"pattern = "*****ba*****ab" # char pattern[] = "ba*****ab"# char pattern[] = "ba * ab"# char pattern[] = "a * ab"if (stringmatch(strr, pattern, len(strr),                               len(pattern))):    print("Yes")else:    print( "No") pattern2 = "a*****ab";if (stringmatch(strr, pattern2, len(strr),                                len(pattern2))):    print("Yes")else:    print( "No") # This code is contributed# by sahilhelangia
// C# program to implement wildcard// pattern matching algorithmusing System; class GFG {     // Function that matches input text    // with given wildcard pattern    static Boolean strmatch(char[] txt, char[] pat,                            int n, int m)    {        // empty pattern can only        // match with empty string.        // Base Case :        if (m == 0)            return (n == 0);         // step-1 :        // initialize markers :        int i = 0, j = 0, index_txt = -1,            index_pat = -1;         while (i < n) {             // For step - (2, 5)            if (j < m && txt[i] == pat[j]) {                i++;                j++;            }             // For step - (3)            else if (j < m && pat[j] == '?') {                i++;                j++;            }             // For step - (4)            else if (j < m && pat[j] == '*') {                index_txt = i;                index_pat = j;                j++;            }             // For step - (5)            else if (index_pat != -1) {                j = index_pat + 1;                i = index_txt + 1;                index_txt++;            }             // For step - (6)            else {                return false;            }        }         // For step - (7)        while (j < m && pat[j] == '*') {            j++;        }         // Final Check        if (j == m) {            return true;        }         return false;    }     // Driver code    public static void Main(String[] args)    {        char[] str = "baaabab".ToCharArray();        char[] pattern = "*****ba*****ab".ToCharArray();        // char pattern[] = "ba*****ab";        // char pattern[] = "ba*ab";        // char pattern[] = "a*ab";         if (strmatch(str, pattern, str.Length,                     pattern.Length))            Console.WriteLine("Yes");        else            Console.WriteLine("No");         char[] pattern2 = "a*****ab".ToCharArray();        if (strmatch(str, pattern2, str.Length,                     pattern2.Length))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by Rajput-Ji
<script>    // Javascript program to implement wildcard    // pattern matching algorithm         // Function that matches input text    // with given wildcard pattern    function strmatch(txt, pat, n, m)    {        // empty pattern can only        // match with empty string.        // Base Case :        if (m == 0)            return (n == 0);           // step-1 :        // initialize markers :        let i = 0, j = 0, index_txt = -1,            index_pat = -1;           while (i < n) {               // For step - (2, 5)            if (j < m && txt[i] == pat[j]) {                i++;                j++;            }               // For step - (3)            else if (j < m && pat[j] == '?') {                i++;                j++;            }               // For step - (4)            else if (j < m && pat[j] == '*') {                index_txt = i;                index_pat = j;                j++;            }               // For step - (5)            else if (index_pat != -1) {                j = index_pat + 1;                i = index_txt + 1;                index_txt++;            }               // For step - (6)            else {                return false;            }        }           // For step - (7)        while (j < m && pat[j] == '*') {            j++;        }           // Final Check        if (j == m) {            return true;        }           return false;    }         let str = "baaabab".split('');    let pattern = "*****ba*****ab".split('');     if (strmatch(str, pattern, str.length, pattern.length))      document.write("Yes" + "</br>");    else      document.write("No" + "</br>");     let pattern2 = "a*****ab".split('');    if (strmatch(str, pattern2, str.length, pattern2.length))      document.write("Yes" + "</br>");    else      document.write("No");     </script>
Yes
No
 
Complexity Analysis: 
Time Complexity: O(m + n), where ‘m’ and ‘n’ are the lengths of text and pattern respectively.
Auxiliary Space: O(1). No use of any data structure for storing values
sahilshelangia
aliassar
Rajput-Ji
bidibaaz123
jishantsingh
rameshtravel07
gabaa406
rkbhola5
Competitive Programming
Dynamic Programming
Pattern Searching
Strings
Strings
Dynamic Programming
Pattern Searching
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Prefix Sum Array - Implementation and Applications in Competitive Programming
Ordered Set and GNU C++ PBDS
Modulo 10^9+7 (1000000007)
What is Competitive Programming and How to Prepare for It?
Bits manipulation (Important tactics)
0-1 Knapsack Problem | DP-10
Largest Sum Contiguous Subarray
Bellman–Ford Algorithm | DP-23
Longest Common Subsequence | DP-4
Floyd Warshall Algorithm | DP-16