From e36f48e6d5397b7f4a1f13e701492c84cc1c1990 Mon Sep 17 00:00:00 2001 From: Lohithgoud <137982071+Lohithgoud@users.noreply.github.com> Date: Wed, 6 Dec 2023 19:09:39 +0530 Subject: [PATCH 1/3] Add files via upload --- STRINGS/MEDIUM/word break problem.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) create mode 100644 STRINGS/MEDIUM/word break problem.md diff --git a/STRINGS/MEDIUM/word break problem.md b/STRINGS/MEDIUM/word break problem.md new file mode 100644 index 00000000..e69de29b From 07e496987396c4114481ac709016f5aeceb8a6fb Mon Sep 17 00:00:00 2001 From: Lohithgoud <137982071+Lohithgoud@users.noreply.github.com> Date: Wed, 6 Dec 2023 19:15:30 +0530 Subject: [PATCH 2/3] word break problem.md --- STRINGS/MEDIUM/word break problem.md | 231 +++++++++++++++++++++++++++ 1 file changed, 231 insertions(+) diff --git a/STRINGS/MEDIUM/word break problem.md b/STRINGS/MEDIUM/word break problem.md index e69de29b..868222c0 100644 --- a/STRINGS/MEDIUM/word break problem.md +++ b/STRINGS/MEDIUM/word break problem.md @@ -0,0 +1,231 @@ +# WORDBREAK PROBLEM + +The Word Break Problem is a common problem in computer science and natural language processing. It involves determining if a given string can be segmented into a space-separated sequence of valid words from a dictionary. + +# INTRODUCTION + +The Word Break Problem is a classic dynamic programming problem in computer science. It involves determining whether a given string can be segmented into a space-separated sequence of one or more words, using a given dictionary of words. In other words, the problem is about breaking a string into valid words from a predefined set of words. + + +# OVERVIEW OF THE WORD BREAK PROBLEM + +To solve the Word Break Problem, dynamic programming techniques are commonly employed. The idea is to break down the problem into smaller subproblems and use the solutions of those subproblems to build the solution for the original problem. By efficiently storing and reusing the computed results, dynamic programming algorithms can provide an optimal solution. + +Here are the key points of the problem: + +**Input:** + +- A string s. +- A dictionary of words wordDict. + +**Output:** + +- True if s can be segmented into words from wordDict, False otherwise. + +### **example:** + +**Input:** + +- s = "applepie" +- wordDict = `["apple", "pie", "pen", "pineapple"]` + +**Output:** + + True + +**Explanation:** + +There are two valid ways to segment "applepie" into words from the dictionary: + +1. "apple pie" - Both "apple" and "pie" are present in the dictionary. +2. "applepie" - Both "app" and "lepie" are not present in the dictionary, but "applepie" itself is present as a single word in the dictionary. + +Therefore, the output is True since at least one valid segmentation exists. + +# CODE + +### PYTHON + +```python +# Copyrights to venkys.io + # For more information, visit https://venkys.io + # Time Complexity: O(2^n). +# The space complexity: O(n). + +def wordBreak(string, words): +# Create a list to store whether the substring up to index i can be segmented + d=[False]*len(string) + + for i in range(len(string)): + for w in words: + if w == string[i-len(w)+1:i+1] and (d[i-len(w)] or i-len(w) == -1): + # Empty string is always breakable + d[i]=True + + # print(d) + return d[-1] + +if __name__=="__main__": + string="applepenapple" + words=["apple","pen"] + print(wordBreak(string,words)) + +# Output: True +``` + +# STEP BY STEP EXPLAINTATION FOR PYTHON + +The provided code is a Python implementation of the Word Break Problem using dynamic programming. It checks whether a given string can be segmented into words from a list of provided words. + +Here's a brief explanation of the code: + +1. The wordBreakfunction takes two parameters: string, which is the input string to be checked, and words, which is a list of words that are allowed in the segmentation. +2. It initializes a boolean list d, where d[i] represents whether the substring of the input string up to index i can be segmented into words from the words list. The d list is initialized with False values, indicating that no segmentation is possible initially. +3. It iterates through the characters of the input string using a loop that goes from 0 to the length of the string. +4. Within the loop, it iterates through the words in the words list. For each word w, it checks if the substring of the input string from i - len(w) + 1 to i matches the word w. If it's a match and d[i - len(w)] is True (indicating that the substring up to i - len(w) can be segmented) or if i - len(w) is -1 (indicating that the word w can start at the beginning of the string), then it sets d[i] to True. +5. After the loops have finished, the d list will have been updated to indicate which parts of the input string can be segmented into words from the words list. +6. The function returns the value at the last index of the d list, d[-1], which represents whether the entire input string can be segmented. +7. In the if __name__=="__main__": block, an example is provided. The input string is "applepenapple," and the list of words contains "apple" and "pen." It prints the result of calling the wordBreak function with these inputs. + +The code uses dynamic programming to efficiently solve the Word Break Problem by building a boolean list d that tracks whether various substrings can be segmented into the provided words. The result is True if the entire input string can be segmented, and False otherwise. In this example, it should print True because "applepenapple" can be segmented into "apple" and "pen." + +### JAVA + +```java +/* Copyrights to venkys.io + For more information, visit https://venkys.io + Time Complexity: O(2^n). + The space complexity: O(n). */ + +import java.util.ArrayList; +import java.util.HashSet; + +public class test{ + + static boolean wordBreak(String s,ArrayList words){ + HashSet set = new HashSet<>(words); + boolean[] dp = new boolean[s.length()+1]; + dp[0]=true; + for(int i=1;i<=s.length();i++){ + for(int j=0;j words=new ArrayList<>(); + words.add("apple"); + words.add("pen"); + if (wordBreak(s, words)) { + System.out.println("The string can be segmented."); + } else { + System.out.println("The string cannot be segmented."); + } + + } +} +``` + +# STEP BY STEP EXPLAINTATION FOR JAVA + +The provided code is a Java program that checks whether a given string can be segmented into words from a given list of words. + +Here is a step-by-step explanation of the code: + +1. The necessary classes, ArrayList and HashSet, are imported from the java.util package. +2. The code defines a public class named test. +3. The wordBreak method is defined, which takes a string s and an ArrayList of strings words as parameters, and returns a boolean value. +4. A HashSet named set is created and initialized with the contents of the words list. This allows for efficient lookup of words. +5. A boolean array dp is created with a length one greater than the length of the input string s. The dp array will store whether a substring of s can be segmented into words from the words list. The initial value of dp[0] is set to true. +6. Two nested for loops iterate over the characters of the input string s. +7. For each pair of indices j and i, the code checks if the substring from index j to i (exclusive) can be found in the set of words. If the substring can be found and dp[j] is true, it means that the substring can be segmented. In that case, dp[i] is set to true and the inner loop is exited. +8. The value of dp[s.length()] is returned, indicating whether the entire string s can be segmented into words from the words list. +9. The main method is defined as the entry point of the program. +10. A string variable s is declared and initialized with the value "applepenapple". An ArrayList of strings named words is created and two words, "apple" and "pen", are added to the list. +11. The wordBreak method is called with the s string and words list as arguments. Depending on the returned value, either "The string can be segmented." or "The string cannot be segmented." is printed. + +Overall, the code checks whether a given string can be segmented into words from a list using dynamic programming. + +### C++ + +```cpp +/* Copy rights to venkys.io +For more information visit https://venkys.io*/ + +#include + +using namespace std; + +// dp[i] will be true if the first i characters of s can be segmented into words in the wordDict +bool wordBreak(std::string s,std::vector& wordDict){ + std::set word_set(wordDict.begin(),wordDict.end()); + int n=s.size(); + std::vectordp(n+1,0); + dp[0]=1; + for(int i=0;i words{ "apple", "pen" }; + if (wordBreak(s, words)) { + std::cout << "The string can be segmented into words." << std::endl; + } else { + std::cout << "The string cannot be segmented into words." << std::endl; + } + + return 0; + +} +``` + +# STEP BY STEP EXPLAINTATION FOR C++ + +The given code is in C++ and it demonstrates the implementation of the word break problem using dynamic programming. The goal of the problem is to determine if a given string can be segmented into words from a given dictionary. + +**Algorithm Implementation** + +The provided code demonstrates the implementation of the word break problem using dynamic programming in C++. The wordBreak function takes a string s and a vector wordDict as input. It initially converts wordDict into a set word_set to facilitate efficient lookup. Subsequently, it initializes a dynamic programming array dp of size n+1, where n represents the length of the string s. + +The function begins by setting dp[0] to 1, indicating that an empty string can be segmented into words. It then proceeds through the string using two nested loops. The outer loop iterates from 0 to n-1, while the inner loop iterates from i+1 to n. + +Within the loops, the function checks if dp[i] is true. If it is, it further checks if the substring s.substr(i, j-i) exists in the word_set. If this condition holds, it sets dp[j] to 1, implying that the substring ending at position j can be segmented into words. + +Following the loops, the function returns the value of dp[n], which determines whether the entire string s can be segmented into words. + +**Main Function** + +The main function defines a sample string s and a vector of words words. It then invokes the wordBreak function with s and words as arguments. Based on the function's return value, it prints a message indicating whether the string can be segmented into words. + +**Conclusion** + +The provided code serves as a practical example of solving the word break problem in C++ using dynamic programming. By efficiently segmenting strings into words, this approach proves valuable in various applications. + +# TIME AND SPACE COMPLEXITY + +The time and space complexity of the provided code for the Word Break Problem can be analyzed as follows: + +- **Time Complexity:** The code consists of two nested loops. The outer loop iterates through the characters of the input string, and the inner loop iterates through the words in the provided word list. Therefore, the time complexity can be approximated as `O(n * m),` where n is the length of the input string and m is the number of words in the word list. +- **Space Complexity:** The code uses a boolean list d to store the intermediate results of whether substrings can be segmented into words. The size of the d list is equal to the length of the input string. Therefore, the space complexity is `O(n),` where n is the length of the input string. + +# REAL-WORLD APPLICATION FOR WORDBREAK PROBLEM + +The Word Break Problem has several real-world applications in computer science and natural language processing. Here are a few examples: + +1. **Spell Checking:** In text editors or word processors, the Word Break Problem can be used to check the spelling of words by breaking the input text into individual words and comparing them against a dictionary of valid words. This helps identify and correct spelling mistakes. +2. **Search Engines**: Search engines use the Word Break Problem to process user queries and match them with relevant documents. By breaking down the query into individual words and matching them against indexed words, search engines can retrieve accurate search results. +3. **Sentiment Analysis:** The Word Break Problem is used in sentiment analysis tasks, where the goal is to determine the sentiment or emotion associated with a given text. By breaking down the text into words and analyzing the sentiment of each word, sentiment analysis models can classify the overall sentiment of the text. +4. **Machine Translation:** In machine translation systems, the Word Break Problem is crucial for breaking down sentences in the source language into individual words and then translating them into the target language. This helps maintain the correct word order and structure during the translation process. +5. **Text Segmentation:** Text segmentation is an important task in natural language processing, where the goal is to divide a given text into meaningful segments, such as sentences or paragraphs. The Word Break Problem can be used to segment the text by breaking it into individual words and then grouping them based on punctuation or other criteria. From 8c7e536e709d897cf0204d5a964dd324053d4f95 Mon Sep 17 00:00:00 2001 From: Lohithgoud <137982071+Lohithgoud@users.noreply.github.com> Date: Mon, 11 Dec 2023 21:54:01 +0530 Subject: [PATCH 3/3] Create REGULAR EXPRESSION MATCHING.md --- STRINGS/HARD/REGULAR EXPRESSION MATCHING.md | 260 ++++++++++++++++++++ 1 file changed, 260 insertions(+) create mode 100644 STRINGS/HARD/REGULAR EXPRESSION MATCHING.md diff --git a/STRINGS/HARD/REGULAR EXPRESSION MATCHING.md b/STRINGS/HARD/REGULAR EXPRESSION MATCHING.md new file mode 100644 index 00000000..43cd22d7 --- /dev/null +++ b/STRINGS/HARD/REGULAR EXPRESSION MATCHING.md @@ -0,0 +1,260 @@ +# INTRODUCTION TO REGULAR EXPRESSION MATCHING + +Regular expression matching (often referred to as regex or regexp matching) is a powerful and flexible way to search, match, and manipulate text based on patterns. A regular expression is a sequence of characters that defines a search pattern. These patterns can include a variety of elements, such as literals, metacharacters, and quantifiers, allowing for complex and flexible text matching.. + +# OVERVIEW OF REGULAR EXPRESSION MATCHING + +The Regular Expression Matching problem involves determining if a given string matches a specified pattern defined by a regular expression. This problem is commonly encountered in string matching, text processing, and pattern recognition tasks. The regular expression specifies a set of rules that the input string must follow for a match to occur. + +Given two strings **S** and **P** where **S** consists of only lowercase English alphabets while **P** consists of lowercase English alphabets as well as special characters ‘**.’** and ‘***’,** the task is to implement a function to test regular expression such that: + +- `'.'` Matches any single character. +- `'*'` Matches zero or more of the preceding element. + +### **Here's an example to illustrate the Regular Expression Matching problem:** + +Let's consider the regular expression **`a*b`**: + +- **`a*`**: Zero or more occurrences of the character 'a'. +- **`b`**: The character 'b'. + +Now, suppose we have the following strings: + +1. Input: **`"b"`** + - Does **`"b"`** match the pattern **`a*b`**? + - No, because there are no 'a' characters before 'b'. +2. Input: **`"aaab"`** + - Does **`"aaab"`** match the pattern **`a*b`**? + - Yes, because there are zero or more 'a' characters followed by 'b'. + - + +This problem can be solved using dynamic programming or recursion with memoization. The idea is to build a table or use memoization to store the results of subproblems, avoiding redundant computations. + +# CODE + + PYTHON + +```python +# Copyrights to venkys.io +# For more information, visit https://venkys.io + +def is_match(s, p): + # Initialize a 2D table to store results of subproblems + dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)] + + # Empty pattern matches empty string + dp[0][0] = True + + # Handle patterns with '*' at the beginning + for j in range(1, len(p) + 1): + if p[j - 1] == '*': + dp[0][j] = dp[0][j - 2] + + # Build the table using dynamic programming + for i in range(1, len(s) + 1): + for j in range(1, len(p) + 1): + if p[j - 1] == s[i - 1] or p[j - 1] == '.': + dp[i][j] = dp[i - 1][j - 1] + elif p[j - 1] == '*': + dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] if s[i - 1] == p[j - 2] or p[j - 2] == '.' else False) + + return dp[len(s)][len(p)] + +# Test cases +print(is_match("aa", "a")) # Output: False +print(is_match("aa", "a*")) # Output: True +print(is_match("ab", ".*")) # Output: True +``` + +# STEP BY STEP EXPLAINTATION + +The provided code is an implementation of regular expression matching using dynamic programming. Here's a step-by-step explanation of how the code works: + +1. The code defines a function is_match that takes two strings s and p as input and returns a boolean indicating whether s matches the pattern p. +2. It initializes a 2D table dp with dimensions (len(s) + 1) x (len(p) + 1) to store the results of subproblems. Each cell dp[i][j] represents whether the substring s[:i] matches the pattern p[:j]. +3. It sets dp[0][0] to True to handle the case of an empty pattern matching an empty string. +4. It handles patterns with '*' at the beginning. For each index j from 1 to the length of p, if p[j-1] is '*', it checks dp[0][j-2] to see if the pattern without the preceding character matches an empty string. This is because '*' matches zero or more occurrences of the preceding element. +5. It builds the table dp using dynamic programming. For each index i from 1 to the length of s and each index j from 1 to the length of p, it considers three cases: + - If p[j-1] is equal to s[i-1] or p[j-1] is '.', it means the current characters match. In this case, dp[i][j] is set to dp[i-1][j-1], which indicates that the current substring matches if the previous substrings also match. + - If p[j-1] is '*', it means the current pattern has a preceding element that can be repeated zero or more times. In this case, dp[i][j] is set to dp[i][j-2] if the preceding element is not used (matches zero occurrences). Otherwise, it checks if s[i-1] matches the preceding element p[j-2] or if p[j-2] is '.'. If either condition is true, it sets dp[i][j] to dp[i-1][j], indicating that the current pattern matches if the preceding element is used and the current substring matches the pattern without the '*' and the preceding element. + - Otherwise, dp[i][j] is set to False, indicating that the current substring does not match the pattern. +6. Finally, it returns dp[len(s)][len(p)], which represents whether the entire string s matches the pattern p. + +The provided code includes some test cases to demonstrate the usage of the is_match function. It tests for different scenarios such as matching a single character, matching zero or more occurrences, and using the '.' metacharacter to match any single character. + +# CODE + +### JAVA + +```java +/*Copyrights to vsdevelopers.io*/ +/*For more programs visit vsdevelopers.io */ +/*Java program for regular expression matching*/ + +public class RegularExpressionMatching { + public static boolean isMatch(String s, String p) { + // Initialize a 2D table to store results of subproblems + boolean[][] dp = new boolean[s.length() + 1][p.length() + 1]; + + // Empty pattern matches empty string + dp[0][0] = true; + + // Handle patterns with '*' at the beginning + for (int j = 1; j <= p.length(); j++) { + if (p.charAt(j - 1) == '*') { + dp[0][j] = dp[0][j - 2]; + } + } + + // Build the table using dynamic programming + for (int i = 1; i <= s.length(); i++) { + for (int j = 1; j <= p.length(); j++) { + if (p.charAt(j - 1) == s.charAt(i - 1) || p.charAt(j - 1) == '.') { + dp[i][j] = dp[i - 1][j - 1]; + } else if (p.charAt(j - 1) == '*') { + dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.')); + } + } + } + + return dp[s.length()][p.length()]; + } + + public static void main(String[] args) { + // Test cases + System.out.println(isMatch("aa", "a")); // Output: false + System.out.println(isMatch("aa", "a*")); // Output: true + System.out.println(isMatch("ab", ".*")); // Output: true + } +} +``` + +# STEP BY STEP EXPLAINTATION + +The provided code is an implementation of regular expression matching using dynamic programming. Here's a step-by-step explanation of how the code works: + +1. Initialize a 2D table, dp, to store the results of subproblems. +2. Set dp[0][0] to true to indicate that an empty pattern matches an empty string. +3. Handle patterns with '*' at the beginning by setting dp[0][j] to dp[0][j-2] for all indices j where p.charAt(j-1) is '*'. +4. Iterate through the strings s and p using nested loops. +5. If the current characters at indices i and j in s and p are the same, or the current character in p is '.', set dp[i][j] to dp[i-1][j-1]. +6. If the current character in p is '*', set dp[i][j] to dp[i][j-2] (ignoring the '*' and the character before it) or (dp[i-1][j] && (s.charAt(i-1) == p.charAt(j-2) || p.charAt(j-2) == '.')) (matching the current character in s with the character before the '*' in p). +7. Return dp[s.length()][p.length()] as the result of the matching. + +The provided code includes test cases to demonstrate the functionality of the regular expression matching algorithm. + +# CODE + +### C++ + +```cpp + +//Copyrights to vsdevelopers.io +//For more programs visit vsdevelopers.io + + +#include +#include + +using namespace std; + +bool isMatch(string s, string p) { + // Initialize a 2D table to store results of subproblems + vector> dp(s.length() + 1, vector(p.length() + 1, false)); + + // Empty pattern matches empty string + dp[0][0] = true; + + // Handle patterns with '*' at the beginning + for (int j = 1; j <= p.length(); j++) { + if (p[j - 1] == '*') { + dp[0][j] = dp[0][j - 2]; + } + } + + // Build the table using dynamic programming + for (int i = 1; i <= s.length(); i++) { + for (int j = 1; j <= p.length(); j++) { + if (p[j - 1] == s[i - 1] || p[j - 1] == '.') { + dp[i][j] = dp[i - 1][j - 1]; + } else if (p[j - 1] == '*') { + dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')); + } + } + } + + return dp[s.length()][p.length()]; +} + +int main() { + // Test cases + cout << isMatch("aa", "a") << endl; // Output: false + cout << isMatch("aa", "a*") << endl; // Output: true + cout << isMatch("ab", ".*") << endl; // Output: true + + return 0; +} + +``` + +## + +# STEP BY STEP EXPLAINTATION + +1. We start by including the necessary header files and declaring the isMatch function which takes two strings, s and p, as parameters and returns a boolean value. +2. Inside the isMatch function, we initialize a 2D table dp to store the results of subproblems. The size of the table is s.length() + 1 rows and p.length() + 1 columns. Each entry is initially set to false. +3. We set dp[0][0] to true since an empty pattern matches an empty string. +4. Next, we handle patterns with '*' at the beginning. For each j from 1 to the length of p, if p[j - 1] is '*', we set dp[0][j] to dp[0][j - 2]. This means that the '*' can match zero occurrences of the preceding character. +5. We build the table using dynamic programming. For each i from 1 to the length of s, and for each j from 1 to the length of p, we consider three cases: + - If p[j - 1] is equal to s[i - 1] or p[j - 1] is '.', we set dp[i][j] to dp[i - 1][j - 1]. This means that the current characters match, and we can move to the next characters. + - If p[j - 1] is '*', we have two subcases: + - The '*' matches zero occurrences of the preceding character, so we set dp[i][j] to dp[i][j - 2]. + - The '*' matches one or more occurrences of the preceding character. In this case, we check if s[i - 1] is equal to p[j - 2] or p[j - 2] is '.' and dp[i - 1][j] is true. If either of these conditions is true, we set dp[i][j] to true. +6. Finally, we return dp[s.length()][p.length()], which represents whether the entire string s matches the pattern p. +7. In the main function, we test the isMatch function with three test cases and print the results. + +The code uses dynamic programming to solve the regular expression matching problem. It considers all possible combinations of characters in s and p and stores the results in the dp table. By the end, it determines whether s matches p based on the values in the table. + +# TIME AND SPACE COMPLEXITY : + +The time complexity is `O(m * n)`, where m is the length of the input string **s**, and n is the length of the pattern string **p**. This is because there is a nested loop iterating through the lengths of both strings. + +1. The outer loop runs for **m + 1** iterations. +2. The inner loop runs for **n + 1** iterations. + +Each iteration involves constant-time operations, so the overall time complexity is O(m * n). + +The space complexity is also `O(m * n)` due to the space used by the 2D table . + +1. The table has dimensions **(m + 1) x (n + 1)**, where **m** is the length of string **s**, and **n** is the length of pattern **p**. +2. Each entry in the table stores a boolean value. + +Therefore, the space complexity is O(m * n). + +# REAL-WORLD APPLICATION FOR REGULAR EXPRESSION MATCHING + +Regular expression matching has numerous real-world applications across various domains due to its ability to define and search for complex patterns in text data. Here are some common real-world applications: + +1. **Text Search and Validation:** + - **Search Engines:** Search engines use regular expressions to match user queries against a vast amount of text data efficiently. + - **Text Editors and IDEs:** Text editors and integrated development environments (IDEs) often provide find-and-replace functionality using regular expressions. +2. **Data Validation and Extraction:** + - **Form Validation:** Regular expressions are commonly used to validate user input in forms, such as email addresses, phone numbers, or ZIP codes. + - **Log Analysis:** Regular expressions help extract specific information from log files, enabling analysis and troubleshooting. +3. **String Manipulation and Parsing:** + - **Data Cleaning:** Regular expressions are employed to clean and preprocess textual data by removing or replacing specific patterns. + - **URL Parsing:** Regular expressions can be used to parse and extract components from URLs, such as extracting the domain or parameters. +4. **Lexical Analysis and Compilers:** + - **Programming Languages:** Regular expressions play a vital role in lexical analysis, where they are used to define tokens in the source code of programming languages. + - **Compiler Construction:** Regular expressions are part of the toolkit used in the construction of compilers for parsing and tokenizing code. +5. **Natural Language Processing (NLP):** + - **Named Entity Recognition:** Regular expressions can be used to define patterns for identifying named entities (e.g., names, locations) in text data. + - **Text Pattern Matching:** In NLP tasks, regular expressions are applied to match specific linguistic patterns or structures. +6. **Network Security:** + - **Intrusion Detection Systems (IDS):** Regular expressions are used to define patterns of known attack signatures or suspicious network activities in security systems. + - **Log Analysis for Security:** Regular expressions aid in extracting relevant information from security logs for analysis and threat detection. +7. **Web Scraping and Data Extraction:** + - **Web Scraping:** Regular expressions are utilized to extract specific data patterns from HTML or other markup languages. + - **Data Extraction from Documents:** Regular expressions can be employed to extract structured information from documents in various formats. +8. **Configuration File Parsing:** + - **Parsing Configuration Files:** Regular expressions are used to parse and extract information from configuration files in software applications.