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*ksforneveropen”, text = “neveropen”
Output: Yes
Explanation:
Replacing ‘*’ with ‘e’, modifies pattern equal to “neveropen”.
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: Below is the idea to solve the problem:
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?))
Check if a given pattern exists in a given string or not using Dynamic Programming
Below is the idea to solve the problem:
Construct a 2D array dp[M+1][N+1] with DP state dp[i][j] denoting if the first i characters of string text match with the first j characters of pattern then assign dp[i][j] as 1 else 0 according to the match of text[i] and pattern[j] and true value of dp[i-1][j-1].
- Initialize a 2D array, dp[M + 1][N + 1], where dp[i][j] check if the substring {text[0], …, text[i]} 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 characters match so fill dp[i + 1][j + 1] = dp[i][j].
- If pattern[j] is ‘.’ then characters match so 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 characters don’t match so 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 notint 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 Codeint main(){ string text = "neveropen"; string pattern = "ge*ksforneveropen"; if (isMatch(text, pattern)) cout << "Yes"; else cout << "No";}// This code is contributed by mohiy kumar 29. |
Java
// Java program for the above approachimport 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 = "neveropen"; String pattern = "ge*ksforneveropen"; 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 notdef 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)) rows, cols = (N + 1, M + 1) dp = [[0]*cols]*rows # 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 Codeif __name__ == "__main__": text = "neveropen" pattern = "ge*ksforneveropen" if (isMatch(text, pattern)): print("Yes") else: print("No")# This code is contributed by AnkThon |
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 = "neveropen"; let pattern = "ge*ksforneveropen"; if (isMatch(text, pattern)) { document.write("Yes"); } else { document.write("No"); }</script> |
C#
// C# program for the above approachusing 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 = "neveropen"; string pattern = "ge*ksforneveropen"; if (isMatch(text, pattern)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by ukasp |
Output
Yes
Time Complexity: O(M * N)
Auxiliary Space: O(M * N)
Efficient Approach : using array instead of 2d matrix to optimize space complexity
In previous code we can se that dp[i][j] is dependent upon dp[i+1][j-1] or dp[i][j-1] so we can assume that dp[i+1] is current row and dp[i] is previous row.
Implementations Steps :
- Initialize a 1D boolean vector ‘dp’ of size M+1, where M is the length of the pattern.
- Base case: dp[0] is true, as an empty pattern matches an empty text.
- Iterate over the characters of the pattern. If the current character is ‘*’, update dp[i+1] to true if dp[i-1] is true.
- Iterate over the characters of the text. Keep a variable ‘prev’ to store the previous value of dp[j]. For each character in the pattern, update dp[j+1] based on the current and previous values of dp[j], dp[j-1], and prev, depending on the character in the pattern.
- Return dp[M] as the final answer. If dp[M] is true, the pattern matches the text; otherwise, it does not.
Implementation :
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 notint 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[j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not vector<bool> dp(M + 1, false); // Base Case dp[0] = true; // Iterate over the characters // of the string pattern for (int i = 0; i < M; i++) { if (pattern[i] == '*' && dp[i - 1]) { // Update dp[i + 1] dp[i + 1] = true; } } // Iterate over the characters // of both the strings for (int i = 0; i < N; i++) { bool prev = dp[0]; dp[0] = false; for (int j = 0; j < M; j++) { bool temp = dp[j+1]; // If current character // in the pattern is '.' if (pattern[j] == '.') { // Update dp[j + 1] dp[j + 1] = prev; } // If current character in // both the strings are equal else if (pattern[j] == text[i]) { // Update dp[j + 1] dp[j + 1] = prev; } // If current character // in the pattern is '*' else if (pattern[j] == '*') { if (pattern[j - 1] != text[i] && pattern[j - 1] != '.') { // Update dp[j + 1] dp[j + 1] = dp[j - 1]; } else { // Update dp[j + 1] dp[j + 1] = (dp[j + 1] or dp[j] or prev); } } else { // Update dp[j + 1] dp[j + 1] = false; } prev = temp; } } // Return dp[M] return dp[M];}// Driver Codeint main(){ string text = "neveropen"; string pattern = "ge*ksforneveropen"; if (isMatch(text, pattern)) cout << "Yes"; else cout << "No";}// this code is contributed by bhardwajji |
Java
// Java program for the above approachimport java.util.*;public class Main { // Function to check if the pattern // consisting of '*', '.' and lowercase // characters matches the text or not static int isMatch(String text, String pattern) { // Base Case if (text.equals("") || pattern.equals("")) return 0; // Stores length of text int N = text.length(); // Stores length of pattern int M = pattern.length(); // dp[j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not boolean[] dp = new boolean[M + 1]; // Base Case dp[0] = true; // Iterate over the characters // of the string pattern for (int i = 0; i < M; i++) { if (pattern.charAt(i) == '*' && dp[i - 1]) { // Update dp[i + 1] dp[i + 1] = true; } } // Iterate over the characters // of both the strings for (int i = 0; i < N; i++) { boolean prev = dp[0]; dp[0] = false; for (int j = 0; j < M; j++) { boolean temp = dp[j + 1]; // If current character // in the pattern is '.' if (pattern.charAt(j) == '.') { // Update dp[j + 1] dp[j + 1] = prev; } // If current character in // both the strings are equal else if (pattern.charAt(j) == text.charAt(i)) { // Update dp[j + 1] dp[j + 1] = prev; } // If current character // in the pattern is '*' else if (pattern.charAt(j) == '*') { if (j > 0 && pattern.charAt(j - 1) != text.charAt(i) && pattern.charAt(j - 1) != '.') { // Update dp[j + 1] dp[j + 1] = dp[j - 1]; } else { // Update dp[j + 1] dp[j + 1] = (dp[j + 1] || dp[j] || prev); } } else { // Update dp[j + 1] dp[j + 1] = false; } prev = temp; } } // Return dp[M] return dp[M] ? 1 : 0; } // Driver Code public static void main(String[] args) { String text = "neveropen"; String pattern = "ge*ksforneveropen"; if (isMatch(text, pattern) == 1) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python program for the above approach# Function to check if the pattern# consisting of '*', '.' and lowercase# characters matches the text or notdef 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[j]: Check if { text[0], .. text[i] } # matches {pattern[0], ... pattern[j]} or not dp = [False] * (M + 1) # Base Case dp[0] = True # Iterate over the characters # of the string pattern for i in range(M): if pattern[i] == '*' and dp[i - 1]: # Update dp[i + 1] dp[i + 1] = True # Iterate over the characters # of both the strings for i in range(N): prev = dp[0] dp[0] = False for j in range(M): temp = dp[j + 1] # If current character # in the pattern is '.' if pattern[j] == '.': # Update dp[j + 1] dp[j + 1] = prev # If current character in # both the strings are equal elif pattern[j] == text[i]: # Update dp[j + 1] dp[j + 1] = prev # If current character # in the pattern is '*' elif pattern[j] == '*': if pattern[j - 1] != text[i] and pattern[j - 1] != '.': # Update dp[j + 1] dp[j + 1] = dp[j - 1] else: # Update dp[j + 1] dp[j + 1] = (dp[j + 1] or dp[j] or prev) else: # Update dp[j + 1] dp[j + 1] = False prev = temp # Return dp[M] return dp[M]# Driver Codeif __name__ == '__main__': text = "neveropen" pattern = "ge*ksforneveropen" if isMatch(text, pattern): print("Yes") else: print("No") |
Javascript
// Function to check if the pattern// consisting of '*', '.' and lowercase// characters matches the text or notfunction isMatch(text, pattern) { // Base Case if (text === "" || pattern === "") { return false; } // Stores length of text let N = text.length; // Stores length of pattern let M = pattern.length; // dp[j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not let dp = new Array(M + 1).fill(false); // Base Case dp[0] = true; // Iterate over the characters // of the string pattern for (let i = 0; i < M; i++) { if (pattern[i] === '*' && dp[i - 1]) { // Update dp[i + 1] dp[i + 1] = true; } } // Iterate over the characters // of both the strings for (let i = 0; i < N; i++) { let prev = dp[0]; dp[0] = false; for (let j = 0; j < M; j++) { let temp = dp[j+1]; // If current character // in the pattern is '.' if (pattern[j] === '.') { // Update dp[j + 1] dp[j + 1] = prev; } // If current character in // both the strings are equal else if (pattern[j] === text[i]) { // Update dp[j + 1] dp[j + 1] = prev; } // If current character // in the pattern is '*' else if (pattern[j] === '*') { if (pattern[j - 1] !== text[i] && pattern[j - 1] !== '.') { // Update dp[j + 1] dp[j + 1] = dp[j - 1]; } else { // Update dp[j + 1] dp[j + 1] = (dp[j + 1] || dp[j] || prev); } } else { // Update dp[j + 1] dp[j + 1] = false; } prev = temp; } } // Return dp[M] return dp[M];}// Driver Codelet text = "neveropen";let pattern = "ge*ksforneveropen";if (isMatch(text, pattern)) { console.log("Yes");} else { console.log("No");} |
C#
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 == "" || pattern == "") { return false; } // Stores length of text int N = text.Length; // Stores length of pattern int M = pattern.Length; // dp[j]: Check if { text[0], .. text[i] } // matches {pattern[0], ... pattern[j]} or not bool[] dp = new bool[M + 1]; // Base Case dp[0] = true; // Iterate over the characters // of the string pattern for (int i = 0; i < M; i++) { if (pattern[i] == '*' && dp[i - 1]) { // Update dp[i + 1] dp[i + 1] = true; } } // Iterate over the characters // of both the strings for (int i = 0; i < N; i++) { bool prev = dp[0]; dp[0] = false; for (int j = 0; j < M; j++) { bool temp = dp[j + 1]; // If current character // in the pattern is '.' if (pattern[j] == '.') { // Update dp[j + 1] dp[j + 1] = prev; } // If current character in // both the strings are equal else if (pattern[j] == text[i]) { // Update dp[j + 1] dp[j + 1] = prev; } // If current character // in the pattern is '*' else if (pattern[j] == '*') { if (pattern[j - 1] != text[i] && pattern[j - 1] != '.') { // Update dp[j + 1] dp[j + 1] = dp[j - 1]; } else { // Update dp[j + 1] dp[j + 1] = (dp[j + 1] || dp[j] || prev); } } else { // Update dp[j + 1] dp[j + 1] = false; } prev = temp; } } // Return dp[M] return dp[M]; } static void Main(string[] args) { string text = "neveropen"; string pattern = "ge*ksforneveropen"; if (isMatch(text, pattern)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }} |
Output
Yes
Time Complexity: O(M * N), where N is the size of the text string and M is the size of the pattern string
Auxiliary Space: O(M)1, space used for storing DP values
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