Wildcard Pattern Matching
Last Updated : 20 Nov, 2024
Given a text txt and a wildcard pattern pat, implement a wildcard pattern matching algorithm that finds if the wildcard pattern is matched with the text. The matching should cover the entire text. The wildcard pattern can include the characters '?' and '*' which denote:
- '?' - matches any single character
- '*' - Matches any sequence of characters (including the empty sequence)
Input: txt = "abcde", pat = "a?c*"
Output: true
Explanation: ? matches with b and * matches with "de".
Input: txt = "baaabab", pat = "a*ab"
Output: false
Explanation: Because in string pattern character 'a' at first position, pattern and text can't be matched.
Input: txt = "abc", pat = "*"
Output: true
Explanation: * matches with whole text "abc".
Using Recursion - O(2^(n+m)) Time and O(n) Space
We start matching the last characters of the both pattern and text. There are three possible cases:
Case 1: The character is ‘*’ . Here two cases arises as follows:
- We consider that '*' is equal to empty substring, and we move to the next character in pattern.
- We match '*' with one or more characters in Text. Here we move to next character in the text.
Case 2: The character is ‘?’ :
As '?' matches with any single character, we move to the next character in both pattern and text.
Case 3: The character in the pattern is not a wildcard:
If current character in Text matches with current character in Pattern, we move to next character in the Pattern and Text. If they do not match, we return false.
C++ // C++ program for wildcard pattern matching using // recursion #include <iostream> #include <string> using namespace std; bool wildCardRec(string& txt, string& pat, int n, int m) { // Empty pattern can match with a empty text only if (m == 0) return (n == 0); // Empty text can match with a pattern consisting // of '*' only. if (n == 0) { for (int i = 0; i < m; i++) if (pat[i] != '*') return false; return true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt[n - 1] == pat[m - 1] || pat[m - 1] == '?') return wildCardRec(txt, pat, n - 1, m - 1); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat[m - 1] == '*') return wildCardRec(txt, pat, n, m - 1) || wildCardRec(txt, pat, n - 1, m); return false; } bool wildCard(string txt, string pat) { int n = txt.size(); int m = pat.size(); return wildCardRec(txt, pat, n, m); } int main() { string txt= "abcde"; string pat = "a*de"; cout << (wildCard(txt, pat) ? "true" : "false"); return 0; } // Java program for wildcard pattern matching using recursion class GfG { static boolean wildCardRec(String txt, String pat, int n, int m) { // Empty pattern can match with an empty text only if (m == 0) return (n == 0); // Empty text can match with a pattern consisting // of '*' only. if (n == 0) { for (int i = 0; i < m; i++) if (pat.charAt(i) != '*') return false; return true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt.charAt(n - 1) == pat.charAt(m - 1) || pat.charAt(m - 1) == '?') return wildCardRec(txt, pat, n - 1, m - 1); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat.charAt(m - 1) == '*') return wildCardRec(txt, pat, n, m - 1) || wildCardRec(txt, pat, n - 1, m); return false; } static boolean wildCard(String txt, String pat) { int n = txt.length(); int m = pat.length(); return wildCardRec(txt, pat, n, m); } public static void main(String[] args) { String txt = "abcde"; String pat = "a*de"; System.out.println(wildCard(txt, pat) ? "true" : "false"); } } # Python program for wildcard pattern matching using # recursion def wildCardRec(txt, pat, n, m): # Empty pattern can match with an empty text only if m == 0: return n == 0 # Empty text can match with a pattern consisting # of '*' only. if n == 0: for i in range(m): if pat[i] != '*': return False return True # Either the characters match or pattern has '?' # move to the next in both text and pattern if txt[n - 1] == pat[m - 1] or pat[m - 1] == '?': return wildCardRec(txt, pat, n - 1, m - 1) # if the current character of pattern is '*' # first case: It matches with zero character # second case: It matches with one or more characters if pat[m - 1] == '*': return wildCardRec(txt, pat, n, m - 1) or \ wildCardRec(txt, pat, n - 1, m) return False def wildCard(txt, pat): n = len(txt) m = len(pat) return wildCardRec(txt, pat, n, m) if __name__ == "__main__": txt = "abcde" pat = "a*de" print("true" if wildCard(txt, pat) else "false") // C# program for wildcard pattern // matching using recursion using System; class GfG { static bool wildCardRec(string txt, string pat, int n, int m) { // Empty pattern can match with an empty text only if (m == 0) return (n == 0); // Empty text can match with a pattern consisting // of '*' only. if (n == 0) { for (int i = 0; i < m; i++) if (pat[i] != '*') return false; return true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt[n - 1] == pat[m - 1] || pat[m - 1] == '?') return wildCardRec(txt, pat, n - 1, m - 1); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat[m - 1] == '*') return wildCardRec(txt, pat, n, m - 1) || wildCardRec(txt, pat, n - 1, m); return false; } static bool wildCard(string txt, string pat) { int n = txt.Length; int m = pat.Length; return wildCardRec(txt, pat, n, m); } static void Main(string[] args) { string txt = "abcde"; string pat = "a*de"; Console.WriteLine(wildCard(txt, pat) ? "true" : "false"); } } // JavaScript program for wildcard pattern matching using // recursion function wildCardRec(txt, pat, n, m) { // Empty pattern can match with an empty text only if (m === 0) return (n === 0); // Empty text can match with a pattern consisting // of '*' only. if (n === 0) { for (let i = 0; i < m; i++) if (pat[i] !== '*') return false; return true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt[n - 1] === pat[m - 1] || pat[m - 1] === '?') return wildCardRec(txt, pat, n - 1, m - 1); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat[m - 1] === '*') return wildCardRec(txt, pat, n, m - 1) || wildCardRec(txt, pat, n - 1, m); return false; } function wildCard(txt, pat) { let n = txt.length; let m = pat.length; return wildCardRec(txt, pat, n, m); } let txt = "abcde"; let pat = "a*de"; console.log(wildCard(txt, pat) ? "true" : "false"); Using Memoization - O(n*m) Time and O(n*m) Space
In this problem, we can observe that the recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: The result of matching a pattern pat of length m with a text txt of length n, i.e., wildCardRec(txt, pat, n, m), depends on the optimal solutions of its subproblems. If the current characters of pat and txt match (or if the character in pat is a '?'), then the solution depends on wildCardRec(txt, pat, n-1, m-1). If the current character in pat is an *, the solution of the problem will depend on the optimal result of wildCardRec(txt, pat, n, m-1) and wildCardRec(txt, pat, n-1, m).
2. Overlapping Subproblems: When using a recursive approach for the wildcard matching problem, we notice that certain subproblems are solved multiple times. For example, when solving wildCardRec(txt, pat, n, m), we may repeatedly compute results for subproblems like wildCardRec(txt, pat, n-1, m-1) or wildCardRec(txt, pat, n, m-1) in different recursive paths.
- Since there are two parameters change in the recursive solution: the current indices of the text txt (ranging from 0 to n) and the pattern pat (ranging from 0 to m). Thus, we create a 2D array memo of size (n+1) by (m+1) for memoization.
- We initialize this array with -1 to indicate that no subproblem has been computed initially.
- We then modify our recursive solution to first check if memo[n][m] is -1; if it is, then only we proceed with further recursive calls.
C++ // C++ program for wildcard pattern // matching using memoization #include <iostream> #include <vector> #include <string> using namespace std; bool wildCardRec(string& txt, string& pat, int n, int m, vector<vector<int>> &memo) { // Empty pattern can match with a empty text only if (m == 0) return (n == 0); // If result for this sub problem has been // already computed, return it if(memo[n][m] != -1) return memo[n][m]; // Empty text can match with a pattern consisting // of '*' only. if (n == 0) { for (int i = 0; i < m; i++) if (pat[i] != '*') return memo[n][m] = false; return memo[n][m] = true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt[n - 1] == pat[m - 1] || pat[m - 1] == '?') return memo[n][m] = wildCardRec(txt, pat, n - 1, m - 1, memo); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat[m - 1] == '*') return memo[n][m] = wildCardRec(txt, pat, n, m - 1, memo) || wildCardRec(txt, pat, n - 1, m, memo); return memo[n][m] = false; } bool wildCard(string txt, string pat) { int n = txt.size(); int m = pat.size(); vector<vector<int>> memo(n+1, vector<int>(m+1, -1)); return wildCardRec(txt, pat, n, m, memo); } int main() { string txt= "abcde"; string pat = "a*de"; cout << (wildCard(txt, pat) ? "true" : "false"); return 0; } // Java program for wildcard pattern matching // using memoization import java.util.Arrays; class GfG { static boolean wildCardRec(String txt, String pat, int n, int m, int[][] memo) { // Empty pattern can match with an empty text only if (m == 0) return (n == 0); // If result for this subproblem has been // already computed, return it if (memo[n][m] != -1) return memo[n][m] == 1; // Empty text can match with a // pattern consisting of '*' only. if (n == 0) { for (int i = 0; i < m; i++) { if (pat.charAt(i) != '*') { memo[n][m] = 0; return false; } } memo[n][m] = 1; return true; } // Either the characters match or pattern has '?' // Move to the next in both text and pattern if (txt.charAt(n - 1) == pat.charAt(m - 1) || pat.charAt(m - 1) == '?') { memo[n][m] = wildCardRec(txt, pat, n - 1, m - 1, memo) ? 1 : 0; return memo[n][m] == 1; } // If the current character of pattern is '*' // First case: It matches with zero character // Second case: It matches with one or more characters if (pat.charAt(m - 1) == '*') { memo[n][m] = (wildCardRec(txt, pat, n, m - 1, memo) || wildCardRec(txt, pat, n - 1, m, memo)) ? 1 : 0; return memo[n][m] == 1; } memo[n][m] = 0; return false; } static boolean wildCard(String txt, String pat) { int n = txt.length(); int m = pat.length(); int[][] memo = new int[n + 1][m + 1]; for (int[] row : memo) Arrays.fill(row, -1); return wildCardRec(txt, pat, n, m, memo); } public static void main(String[] args) { String txt = "abcde"; String pat = "a*de"; System.out.println(wildCard(txt, pat) ? "true" : "false"); } } # Python program for wildcard pattern # matching using memoization def wildCardRec(txt, pat, n, m, memo): # Empty pattern can match with an empty text only if m == 0: return n == 0 # If result for this subproblem has been # already computed, return it if memo[n][m] != -1: return memo[n][m] # Empty text can match with a pattern consisting # of '*' only. if n == 0: for i in range(m): if pat[i] != '*': memo[n][m] = False return False memo[n][m] = True return True # Either the characters match or pattern has '?' # move to the next in both text and pattern if txt[n - 1] == pat[m - 1] or pat[m - 1] == '?': memo[n][m] = wildCardRec(txt, pat, n - 1, m - 1, memo) return memo[n][m] # if the current character of pattern is '*' # first case: It matches with zero character # second case: It matches with one or more characters if pat[m - 1] == '*': memo[n][m] = wildCardRec(txt, pat, n, m - 1, memo) \ or wildCardRec(txt, pat, n - 1, m, memo) return memo[n][m] memo[n][m] = False return False def wildCard(txt, pat): n = len(txt) m = len(pat) memo = [[-1 for _ in range(m + 1)] for _ in range(n + 1)] return wildCardRec(txt, pat, n, m, memo) if __name__ == "__main__": txt = "abcde" pat = "a*de" print("true" if wildCard(txt, pat) else "false") // C# program for wildcard pattern matching // using memoization using System; class GfG { static bool WildCardRec(string txt, string pat, int n, int m, int[,] memo) { // Empty pattern can match with an empty // text only if (m == 0) return (n == 0); // If result for this subproblem has been // already computed, return it if (memo[n, m] != -1) return memo[n, m] == 1; // Empty text can match with a pattern // consisting of '*' only. if (n == 0) { for (int i = 0; i < m; i++) { if (pat[i] != '*') { memo[n, m] = 0; return false; } } memo[n, m] = 1; return true; } // Either the characters match or pattern has '?' // Move to the next in both text and pattern if (txt[n - 1] == pat[m - 1] || pat[m - 1] == '?') { memo[n, m] = WildCardRec(txt, pat, n - 1, m - 1, memo) ? 1 : 0; return memo[n, m] == 1; } // If the current character of pattern is '*' // First case: It matches with zero character // Second case: It matches with one or more characters if (pat[m - 1] == '*') { memo[n, m] = (WildCardRec(txt, pat, n, m - 1, memo) || WildCardRec(txt, pat, n - 1, m, memo)) ? 1 : 0; return memo[n, m] == 1; } memo[n, m] = 0; return false; } static bool WildCard(string txt, string pat) { int n = txt.Length; int m = pat.Length; int[,] memo = new int[n + 1, m + 1]; // Initialize memo array with -1 for (int i = 0; i <= n; i++) { for (int j = 0; j <= m; j++) memo[i, j] = -1; } return WildCardRec(txt, pat, n, m, memo); } static void Main(string[] args) { string txt = "abcde"; string pat = "a*de"; Console.WriteLine(WildCard(txt, pat) ? "true" : "false"); } } // JavaScript program for wildcard pattern // matching using memoization function wildCardRec(txt, pat, n, m, memo) { // Empty pattern can match with an empty text only if (m === 0) return n === 0; // If result for this subproblem has been // already computed, return it if (memo[n][m] !== -1) return memo[n][m]; // Empty text can match with a pattern consisting // of '*' only. if (n === 0) { for (let i = 0; i < m; i++) if (pat[i] !== '*') return memo[n][m] = false; return memo[n][m] = true; } // Either the characters match or pattern has '?' // move to the next in both text and pattern if (txt[n - 1] === pat[m - 1] || pat[m - 1] === '?') return memo[n][m] = wildCardRec(txt, pat, n - 1, m - 1, memo); // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more characters if (pat[m - 1] === '*') return memo[n][m] = wildCardRec(txt, pat, n, m - 1, memo) || wildCardRec(txt, pat, n - 1, m, memo); return memo[n][m] = false; } function wildCard(txt, pat) { let n = txt.length; let m = pat.length; let memo = Array.from(Array(n + 1), () => Array(m + 1).fill(-1)); return wildCardRec(txt, pat, n, m, memo); } const txt = "abcde"; const pat = "a*de"; console.log(wildCard(txt, pat) ? "true" : "false"); Using Bottom-Up DP (Tabulation) - O(n*m) Time and O(n*m) Space
The approach is similar to the previous one; however, instead of solving the problem recursively, we iteratively build the solution using a bottom-up manner. We maintain a dp[][] table such that dp[i][j] stores whether the pattern pat[0...j-1] matches with the text txt[0...i-1].
C++ // C++ program for wild card matching // using tabulation #include <iostream> #include <string> #include <vector> using namespace std; bool wildCard(string& txt, string& pat) { int n = txt.size(); int m = pat.size(); // dp[i][j] will be true if txt[0..i-1] matches pat[0..j-1] vector<vector<bool>> dp(n + 1, vector<bool>(m + 1, false)); // Empty pattern matches with empty string dp[0][0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat[j - 1] == '*') dp[0][j] = dp[0][j - 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat[j - 1] == txt[i - 1] || pat[j - 1] == '?') { // Either the characters match or pattern has '?' // result will be same as previous state dp[i][j] = dp[i - 1][j - 1]; } else if (pat[j - 1] == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more dp[i][j] = dp[i][j - 1] || dp[i - 1][j]; } } } return dp[n][m]; } int main() { string txt = "abcde"; string pat = "a*de"; cout << (wildCard(txt, pat) ? "true" : "false") << endl; return 0; } // Java program for wild card matching using tabulation import java.util.Arrays; class GfG { static boolean wildCard(String txt, String pat) { int n = txt.length(); int m = pat.length(); // dp[i][j] will be true if txt[0..i-1] matches pat[0..j-1] boolean[][] dp = new boolean[n + 1][m + 1]; // Empty pattern matches with empty string dp[0][0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat.charAt(j - 1) == '*') dp[0][j] = dp[0][j - 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat.charAt(j - 1) == txt.charAt(i - 1) || pat.charAt(j - 1) == '?') { // Either the characters match or pattern has '?' // result will be same as previous state dp[i][j] = dp[i - 1][j - 1]; } else if (pat.charAt(j - 1) == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more dp[i][j] = dp[i][j - 1] || dp[i - 1][j]; } } } return dp[n][m]; } public static void main(String[] args) { String txt = "abcde"; String pat = "a*de"; System.out.println(wildCard(txt, pat) ? "true" : "false"); } } # Python program for wild card matching using tabulation def wildCard(txt, pat): n = len(txt) m = len(pat) # dp[i][j] will be True if txt[0..i-1] matches pat[0..j-1] dp = [[False] * (m + 1) for _ in range(n + 1)] # Empty pattern matches with empty string dp[0][0] = True # Handle patterns with '*' at the beginning for j in range(1, m + 1): if pat[j - 1] == '*': dp[0][j] = dp[0][j - 1] for i in range(1, n + 1): for j in range(1, m + 1): if pat[j - 1] == txt[i - 1] or pat[j - 1] == '?': # Either the characters match or pattern has '?' # result will be same as previous state dp[i][j] = dp[i - 1][j - 1] elif pat[j - 1] == '*': # if the current character of pattern is '*' # first case: It matches with zero character # second case: It matches with one or more dp[i][j] = dp[i][j - 1] or dp[i - 1][j] return dp[n][m] if __name__ == "__main__": txt = "abcde" pat = "a*de" print("true" if wildCard(txt, pat) else "false") // C# program for wild card matching using tabulation using System; class GfG { static bool WildCard(string txt, string pat) { int n = txt.Length; int m = pat.Length; // dp[i, j] will be true if txt[0..i-1] matches pat[0..j-1] bool[,] dp = new bool[n + 1, m + 1]; // Empty pattern matches with empty string dp[0, 0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat[j - 1] == '*') dp[0, j] = dp[0, j - 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat[j - 1] == txt[i - 1] || pat[j - 1] == '?') { // Either the characters match or pattern has '?' // result will be same as previous state dp[i, j] = dp[i - 1, j - 1]; } else if (pat[j - 1] == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more dp[i, j] = dp[i, j - 1] || dp[i - 1, j]; } } } return dp[n, m]; } static void Main(string[] args) { string txt = "abcde"; string pat = "a*de"; Console.WriteLine(WildCard(txt, pat) ? "true" : "false"); } } // JavaScript program for wild card matching using tabulation function wildCard(txt, pat) { const n = txt.length; const m = pat.length; // dp[i][j] will be true if txt[0..i-1] matches pat[0..j-1] const dp = Array.from({ length: n + 1 }, () => Array(m + 1).fill(false)); // Empty pattern matches with empty string dp[0][0] = true; // Handle patterns with '*' at the beginning for (let j = 1; j <= m; j++) if (pat[j - 1] === '*') dp[0][j] = dp[0][j - 1]; for (let i = 1; i <= n; i++) { for (let j = 1; j <= m; j++) { if (pat[j - 1] === txt[i - 1] || pat[j - 1] === '?') { // Either the characters match or pattern has '?' // result will be same as previous state dp[i][j] = dp[i - 1][j - 1]; } else if (pat[j - 1] === '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more dp[i][j] = dp[i][j - 1] || dp[i - 1][j]; } } } return dp[n][m]; } const txt = "abcde"; const pat = "a*de"; console.log(wildCard(txt, pat) ? "true" : "false"); Using Space Optimized DP - O(n*m) Time and O(m) Space
If we take a closer look at the above solution, we can notice that we use only row entries, previous and current one. Therefore we can optimize the space by storing only rows.
C++ // C++ program for wild card matching using // space optimized dp #include <iostream> #include <string> #include <vector> using namespace std; bool wildCard(string txt, string pat) { int n = txt.size(); int m = pat.size(); vector<bool> prev(m+1, false); vector<bool> curr(m+1, false); // Empty pattern matches with empty string prev[0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat[j - 1] == '*') prev[j] = prev[j-1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat[j - 1] == txt[i - 1] || pat[j - 1] == '?') { // Either the characters match or pattern has '?' // result will be same as previous state curr[j] = prev[j - 1]; } else if (pat[j - 1] == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more curr[j] = curr[j - 1] || prev[j]; } } prev = curr; } return prev[m]; } int main() { string txt = "abcde"; string pat = "a*de"; cout << (wildCard(txt, pat) ? "true" : "false") << endl; return 0; } // Java program for wild card matching using space optimized // dp class GfG { static boolean wildCard(String txt, String pat) { int n = txt.length(); int m = pat.length(); boolean[] prev = new boolean[m + 1]; boolean[] curr = new boolean[m + 1]; // Empty pattern matches with empty string prev[0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat.charAt(j - 1) == '*') prev[j] = prev[j - 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat.charAt(j - 1) == txt.charAt(i - 1) || pat.charAt(j - 1) == '?') { // Either the characters match or pattern has '?' // result will be same as previous state curr[j] = prev[j - 1]; } else if (pat.charAt(j - 1) == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more curr[j] = curr[j - 1] || prev[j]; } else { curr[j] = false; } } // Copy current row to previous row System.arraycopy(curr, 0, prev, 0, m + 1); } return prev[m]; } public static void main(String[] args) { String txt = "abcde"; String pat = "a*de"; System.out.println(wildCard(txt, pat) ? "true" : "false"); } } # Python program for wild card matching using space optimized # DP def wildCard(txt, pat): n = len(txt) m = len(pat) prev = [False] * (m + 1) curr = [False] * (m + 1) # Empty pattern matches with empty string prev[0] = True # Handle patterns with '*' at the beginning for j in range(1, m + 1): if pat[j - 1] == '*': prev[j] = prev[j - 1] for i in range(1, n + 1): for j in range(1, m + 1): if pat[j - 1] == txt[i - 1] or pat[j - 1] == '?': # Either the characters match or pattern has '?' # result will be same as previous state curr[j] = prev[j - 1] elif pat[j - 1] == '*': # if the current character of pattern is '*' # first case: It matches with zero character # second case: It matches with one or more curr[j] = curr[j - 1] or prev[j] else: curr[j] = False # Copy current row to previous row prev = curr[:] return prev[m] txt = "abcde" pat = "a*de" print("true" if wildCard(txt, pat) else "false") // C# program for wild card matching using Space optimized DP using System; class GfG { static bool WildCard(string txt, string pat) { int n = txt.Length; int m = pat.Length; bool[] prev = new bool[m + 1]; bool[] curr = new bool[m + 1]; // Empty pattern matches with empty string prev[0] = true; // Handle patterns with '*' at the beginning for (int j = 1; j <= m; j++) if (pat[j - 1] == '*') prev[j] = prev[j - 1]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (pat[j - 1] == txt[i - 1] || pat[j - 1] == '?') { // Either the characters match or pattern has '?' // result will be same as previous state curr[j] = prev[j - 1]; } else if (pat[j - 1] == '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more curr[j] = curr[j - 1] || prev[j]; } else { curr[j] = false; } } // Copy current row to previous row Array.Copy(curr, prev, m + 1); } return prev[m]; } static void Main(string[] args) { string txt = "abcde"; string pat = "a*de"; Console.WriteLine(WildCard(txt, pat) ? "true" : "false"); } } // JavaScript program for wild card matching using space optimized // DP function wildCard(txt, pat) { const n = txt.length; const m = pat.length; let prev = new Array(m + 1).fill(false); let curr = new Array(m + 1).fill(false); // Empty pattern matches with empty string prev[0] = true; // Handle patterns with '*' at the beginning for (let j = 1; j <= m; j++) { if (pat[j - 1] === '*') prev[j] = prev[j - 1]; } for (let i = 1; i <= n; i++) { for (let j = 1; j <= m; j++) { if (pat[j - 1] === txt[i - 1] || pat[j - 1] === '?') { // Either the characters match or pattern has '?' // result will be same as previous state curr[j] = prev[j - 1]; } else if (pat[j - 1] === '*') { // if the current character of pattern is '*' // first case: It matches with zero character // second case: It matches with one or more curr[j] = curr[j - 1] || prev[j]; } else { curr[j] = false; } } // Copy current row to previous row prev = [...curr]; } return prev[m]; } const txt = "abcde"; const pat = "a*de"; console.log(wildCard(txt, pat) ? "true" : "false"); Simple Traversal Solution - O(n) Time and O(1) Space
At first, we initialize two pointers i and j to the beginning of the text and the pattern, respectively. We also initialize two variables startIndex and match to -1 and 0, respectively. startIndex will keep track of the position of the last '*' character in the pattern, and match will keep track of the position in the text where the last proper match started.
We then loop through the text until we reach the end or find a character in the pattern that doesn't match the corresponding character in the text. If the current characters match, we simply move to the next characters in both the pattern and the text. Ifnd if the pattern has a '?' , we simply move to the next characters in both the pattern and the text. If the pattern has a '*' character, then we mark the current position in the pattern and the text as a proper match by setting startIndex to the current position in the pattern and its match to the current position in the text. If there was no match and no '*' character, then we understand we need to go through a different route henceforth, we backtrack to the last '*' character position and try a different match by setting j to startIndex + 1, match to match + 1, and i to match.
Once we have looped over the text, we consume any remaining '*' characters in the pattern, and if we have reached the end of both the pattern and the text, the pattern matches the text.
C++ // C++ program for wild card matching using single // traversal #include <iostream> using namespace std; bool wildCard(string txt, string pat) { int n = txt.length(); int m = pat.length(); int i = 0, j = 0, startIndex = -1, match = 0; while (i < n) { // Characters match or '?' in pattern matches // any character. if (j < m && (pat[j] == '?' || pat[j] == txt[i])) { i++; j++; } else if (j < m && pat[j] == '*') { // Wildcard character '*', mark the current // position in the pattern and the text as a // proper match. startIndex = j; match = i; j++; } else if (startIndex != -1) { // No match, but a previous wildcard was found. // Backtrack to the last '*' character position // and try for a different match. j = startIndex + 1; match++; i = match; } else { // If none of the above cases comply, the // pattern does not match. return false; } } // Consume any remaining '*' characters in the given // pattern. while (j < m && pat[j] == '*') { j++; } // If we have reached the end of both the pattern and // the text, the pattern matches the text. return j == m; } int main() { string txt = "baaabab"; string pat = "*****ba*****ab"; cout << (wildCard(txt, pat) ? "true" : "false"); } // Java program for wild card matching using single // traversal class GfG { static boolean wildCard(String txt, String pat) { int n = txt.length(); int m = pat.length(); int i = 0, j = 0, startIndex = -1, match = 0; while (i < n) { // Characters match or '?' in pattern matches // any character. if (j < m && (pat.charAt(j) == '?' || pat.charAt(j) == txt.charAt(i))) { i++; j++; } else if (j < m && pat.charAt(j) == '*') { // Wildcard character '*', mark the current // position in the pattern and the text as a // proper match. startIndex = j; match = i; j++; } else if (startIndex != -1) { // No match, but a previous wildcard was found. // Backtrack to the last '*' character position // and try for a different match. j = startIndex + 1; match++; i = match; } else { // If none of the above cases comply, the // pattern does not match. return false; } } // Consume any remaining '*' characters in the given // pattern. while (j < m && pat.charAt(j) == '*') { j++; } // If we have reached the end of both the pattern and // the text, the pattern matches the text. return j == m; } public static void main(String[] args) { String txt = "baaabab"; String pat = "*****ba*****ab"; System.out.println(wildCard(txt, pat) ? "true" : "false"); } } # Python program for wild card matching using single # traversal def wildCard(txt, pat): n = len(txt) m = len(pat) i = 0 j = 0 startIndex = -1 match = 0 while i < n: # Characters match or '?' in pattern matches # any character. if j < m and (pat[j] == '?' or pat[j] == txt[i]): i += 1 j += 1 elif j < m and pat[j] == '*': # Wildcard character '*', mark the current # position in the pattern and the text as a # proper match. startIndex = j match = i j += 1 elif startIndex != -1: # No match, but a previous wildcard was found. # Backtrack to the last '*' character position # and try for a different match. j = startIndex + 1 match += 1 i = match else: # If none of the above cases comply, the # pattern does not match. return False # Consume any remaining '*' characters in the given # pattern. while j < m and pat[j] == '*': j += 1 # If we have reached the end of both the pattern and # the text, the pattern matches the text. return j == m if __name__ == "__main__": txt = "baaabab" pat = "*****ba*****ab" print("true" if wildCard(txt, pat) else "false") // C# program for wild card matching using single // traversal using System; class GfG { static bool WildCard(string txt, string pat) { int n = txt.Length; int m = pat.Length; int i = 0, j = 0, startIndex = -1, match = 0; while (i < n) { // Characters match or '?' in pattern matches // any character. if (j < m && (pat[j] == '?' || pat[j] == txt[i])) { i++; j++; } else if (j < m && pat[j] == '*') { // Wildcard character '*', mark the current // position in the pattern and the text as a // proper match. startIndex = j; match = i; j++; } else if (startIndex != -1) { // No match, but a previous wildcard was found. // Backtrack to the last '*' character position // and try for a different match. j = startIndex + 1; match++; i = match; } else { // If none of the above cases comply, the // pattern does not match. return false; } } // Consume any remaining '*' characters in the given // pattern. while (j < m && pat[j] == '*') { j++; } // If we have reached the end of both the pattern and // the text, the pattern matches the text. return j == m; } static void Main(string[] args) { string txt = "baaabab"; string pat = "*****ba*****ab"; Console.WriteLine(WildCard(txt, pat) ? "true" : "false"); } } // JavaScript program for wild card matching using single // traversal function wildCard(txt, pat) { let n = txt.length; let m = pat.length; let i = 0, j = 0, startIndex = -1, match = 0; while (i < n) { // Characters match or '?' in pattern matches // any character. if (j < m && (pat[j] === '?' || pat[j] === txt[i])) { i++; j++; } else if (j < m && pat[j] === '*') { // Wildcard character '*', mark the current // position in the pattern and the text as a // proper match. startIndex = j; match = i; j++; } else if (startIndex !== -1) { // No match, but a previous wildcard was found. // Backtrack to the last '*' character position // and try for a different match. j = startIndex + 1; match++; i = match; } else { // If none of the above cases comply, the // pattern does not match. return false; } } // Consume any remaining '*' characters in the given // pattern. while (j < m && pat[j] === '*') { j++; } // If we have reached the end of both the pattern and // the text, the pattern matches the text. return j === m; } let txt = "baaabab"; let pat = "*****ba*****ab"; console.log(wildCard(txt, pat) ? 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