Check if a given pattern exists in a given string or not
Given two strings text and pattern of length M and N respectively, the task is to check if the pattern matches the text or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Note: pattern can include the characters ‘*’ and ‘•’
- ‘*’ matches zero or more occurrences of character right before the current character
- ‘•’ matches any signal character.
Examples:
Input: pattern = “ge*ksforgeeks”, text = “geeksforgeeks”
Output: Yes
Explanation:
Replacing * with ‘e’, modifies pattern equal to “geeksforgeeks”.
Therefore, the required output is Yes.Input: pattern = “ab*d”, text = “abcds”
Output: No
Explanation: The given pattern cannot be matched with the text.
Naive Approach: The simplest approach to solve this problem is to iterate over the characters of the both the strings using recursion. If current character is ‘.’, replace current character to any character and recur for the remaining pattern and text string. Otherwise, if the current character is ‘*’, recur for the remaining text and check if it matches the rest of the pattern or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Time Complexity: O((M + N) * 2(M + N / 2​))
Auxiliary Space: O((M + N) * 2(M + N / 2​))
Efficient Approach: The above approach can be optimized using Dynamic Programming. Following are the recurrence relation:
- Initialize a 2D array, dp[M + 1][N + 1], where dp[i][j] check if the substring { text[0], …, text[i] } is matches with the substring { pattern[0], … pattern[j] } or not.
- Iterate over the characters of the both the strings and fill the dp[][] array based on the following recurrence relation:
- If text[i] and pattern[j] are the same then fill dp[i + 1][j + 1] = dp[i][j].
- If pattern[j] is ‘.’ then fill dp[i + 1][j + 1] = dp[i][j].
- If pattern[j] is ‘*’ then check the following conditions:
- If text[i] is not equal to pattern[j – 1] and pattern[j – 1] is not equal to ‘.’, then fill dp[i + 1][j + 1] = dp[i + 1][j – 1].
- Otherwise, fill dp[i + 1][j + 1] = (dp[i + 1][j] || dp[i][j + 1] || dp[i + 1][j – 1]).
- Finally, print the value of dp[M][N].
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if the pattern // consisting of '*', '.' and lowercase // characters matches the text or not int isMatch(string text, string pattern) { // Base Case if (text == "" or pattern == "" ) return false ; // Stores length of text int N = text.size(); // Stores length of pattern int M = pattern.size(); // dp[i][j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not vector<vector< bool >> dp(N + 1, vector< bool >(M + 1, false )); // Base Case dp[0][0] = true ; // Iterate over the characters // of the string pattern for ( int i = 0; i < M; i++) { if (pattern[i] == '*' && dp[0][i - 1]) { // Update dp[0][i + 1] dp[0][i + 1] = true ; } } // Iterate over the characters // of both the strings for ( int i = 0; i < N; i++) { for ( int j = 0; j < M; j++) { // If current character // in the pattern is '.' if (pattern[j] == '.' ) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i][j]; } // If current character in // both the strings are equal if (pattern[j] == text[i]) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i][j]; } // If current character // in the pattern is '*' if (pattern[j] == '*' ) { if (pattern[j - 1] != text[i] && pattern[j - 1] != '.' ) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i + 1][j - 1]; } else { // Update dp[i+1][j+1] dp[i + 1][j + 1] = (dp[i + 1][j] or dp[i][j + 1] or dp[i + 1][j - 1]); } } } } // Return dp[M][N] return dp[N][M]; } // Driver Code int main() { string text = "geeksforgeeks" ; string pattern = "ge*ksforgeeks" ; if (isMatch(text, pattern)) cout<< "Yes" ; else cout<< "No" ; } // This code is contributed by mohiy kumar 29. |
Java
// Java program for the above approach import java.io.*; class GFG { // Function to check if the pattern // consisting of '*', '.' and lowercase // characters matches the text or not static boolean isMatch(String text, String pattern) { // Base Case if (text == null || pattern == null ) { return false ; } // Stores length of text int N = text.length(); // Stores length of pattern int M = pattern.length(); // dp[i][j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not boolean [][] dp = new boolean [N + 1 ][M + 1 ]; // Base Case dp[ 0 ][ 0 ] = true ; // Iterate over the characters // of the string pattern for ( int i = 0 ; i < M; i++) { if (pattern.charAt(i) == '*' && dp[ 0 ][i - 1 ]) { // Update dp[0][i + 1] dp[ 0 ][i + 1 ] = true ; } } // Iterate over the characters // of both the strings for ( int i = 0 ; i < N; i++) { for ( int j = 0 ; j < M; j++) { // If current character // in the pattern is '.' if (pattern.charAt(j) == '.' ) { // Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i][j]; } // If current character in // both the strings are equal if (pattern.charAt(j) == text.charAt(i)) { // Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i][j]; } // If current character // in the pattern is '*' if (pattern.charAt(j) == '*' ) { if (pattern.charAt(j - 1 ) != text.charAt(i) && pattern.charAt(j - 1 ) != '.' ) { // Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i + 1 ][j - 1 ]; } else { // Update dp[i+1][j+1] dp[i + 1 ][j + 1 ] = (dp[i + 1 ][j] || dp[i][j + 1 ] || dp[i + 1 ][j - 1 ]); } } } } // Return dp[M][N] return dp[N][M]; } // Driver Code public static void main(String[] args) { String text = "geeksforgeeks" ; String pattern = "ge*ksforgeeks" ; if (isMatch(text, pattern)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } |
Python3
# Python3 program for the above approach import numpy as np # Function to check if the pattern # consisting of '*', '.' and lowercase # characters matches the text or not def isMatch(text, pattern): # Base Case if (text = = " " or pattern == " "): return False # Stores length of text N = len (text) # Stores length of pattern M = len (pattern) # dp[i][j]: Check if { text[0], .. text[i] } # matches {pattern[0], ... pattern[j]} or not dp = np.zeros((N + 1 , M + 1 )) # Base Case dp[ 0 ][ 0 ] = True # Iterate over the characters # of the string pattern for i in range (M): if (pattern[i] = = '*' and dp[ 0 ][i - 1 ]): # Update dp[0][i + 1] dp[ 0 ][i + 1 ] = True # Iterate over the characters # of both the strings for i in range (N): for j in range (M): # If current character # in the pattern is '.' if (pattern[j] = = '.' ): # Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i][j] # If current character in # both the strings are equal if (pattern[j] = = text[i]): # Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i][j] # If current character # in the pattern is '*' if (pattern[j] = = '*' ): if (pattern[j - 1 ] ! = text[i] and pattern[j - 1 ] ! = '.' ): # Update dp[i + 1][j + 1] dp[i + 1 ][j + 1 ] = dp[i + 1 ][j - 1 ] else : # Update dp[i+1][j+1] dp[i + 1 ][j + 1 ] = (dp[i + 1 ][j] or dp[i][j + 1 ] or dp[i + 1 ][j - 1 ]) # Return dp[M][N] return dp[N][M] # Driver Code if __name__ = = "__main__" : text = "geeksforgeeks" pattern = "ge*ksforgeeks" if (isMatch(text, pattern)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by AnkThon |
C#
// C# program for the above approach using System; class GFG{ // Function to check if the pattern // consisting of '*', '.' and lowercase // characters matches the text or not static bool isMatch( string text, string pattern) { // Base Case if (text == null || pattern == null ) { return false ; } // Stores length of text int N = text.Length; // Stores length of pattern int M = pattern.Length; // dp[i][j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not bool [,] dp = new bool [N + 1, M + 1]; // Base Case dp[0, 0] = true ; // Iterate over the characters // of the string pattern for ( int i = 0; i < M; i++) { if (pattern[i] == '*' && dp[0, i - 1]) { // Update dp[0][i + 1] dp[0, i + 1] = true ; } } // Iterate over the characters // of both the strings for ( int i = 0; i < N; i++) { for ( int j = 0; j < M; j++) { // If current character // in the pattern is '.' if (pattern[j] == '.' ) { // Update dp[i + 1][j + 1] dp[i + 1, j + 1] = dp[i, j]; } // If current character in // both the strings are equal if (pattern[j] == text[i]) { // Update dp[i + 1][j + 1] dp[i + 1, j + 1] = dp[i, j]; } // If current character // in the pattern is '*' if (pattern[j] == '*' ) { if (pattern[j - 1] != text[i] && pattern[j - 1] != '.' ) { // Update dp[i + 1][j + 1] dp[i + 1, j + 1] = dp[i + 1, j - 1]; } else { // Update dp[i+1][j+1] dp[i + 1, j + 1] = (dp[i + 1, j] || dp[i, j + 1] || dp[i + 1, j - 1]); } } } } // Return dp[M][N] return dp[N, M]; } // Driver Code public static void Main() { string text = "geeksforgeeks" ; string pattern = "ge*ksforgeeks" ; if (isMatch(text, pattern)) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } // This code is contributed by ukasp |
Javascript
<script> // JavaScript program for the above approach // Function to check if the pattern // consisting of '*', '.' and lowercase // characters matches the text or not function isMatch(text, pattern) { // Base Case if (text == null || pattern == null ) { return false ; } // Stores length of text let N = text.length; // Stores length of pattern let M = pattern.length; // dp[i][j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not let dp = new Array(N + 1); for (let i = 0; i <= N; i++) { dp[i] = new Array(M + 1); for (let j = 0; j <= M; j++) { dp[i][j] = false ; } } // Base Case dp[0][0] = true ; // Iterate over the characters // of the string pattern for (let i = 0; i < M; i++) { if (pattern[i] == '*' && dp[0][i - 1]) { // Update dp[0][i + 1] dp[0][i + 1] = true ; } } // Iterate over the characters // of both the strings for (let i = 0; i < N; i++) { for (let j = 0; j < M; j++) { // If current character // in the pattern is '.' if (pattern[j] == '.' ) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i][j]; } // If current character in // both the strings are equal if (pattern[j] == text[i]) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i][j]; } // If current character // in the pattern is '*' if (pattern[j] == '*' ) { if (pattern[j - 1] != text[i] && pattern[j - 1] != '.' ) { // Update dp[i + 1][j + 1] dp[i + 1][j + 1] = dp[i + 1][j - 1]; } else { // Update dp[i+1][j+1] dp[i + 1][j + 1] = (dp[i + 1][j] || dp[i][j + 1] || dp[i + 1][j - 1]); } } } } // Return dp[M][N] return dp[N][M]; } let text = "geeksforgeeks" ; let pattern = "ge*ksforgeeks" ; if (isMatch(text, pattern)) { document.write( "Yes" ); } else { document.write( "No" ); } </script> |
Yes
Time Complexity: O(M * N)
Auxiliary Space: O(M * N)