test: RegexMatchingTest (#5403)

* test: RegexMatchingTest

* checkstyle: fix formatting

---------

Co-authored-by: alxkm <alx@alx.com>
Co-authored-by: Andrii Siriak <siryaka@gmail.com>
This commit is contained in:
Alex Klymenko 2024-08-26 16:22:39 +02:00 committed by GitHub
parent 4374a50fd7
commit 7674a84f5b
No known key found for this signature in database
GPG Key ID: B5690EEEBB952194
2 changed files with 99 additions and 23 deletions

View File

@ -6,17 +6,28 @@ package com.thealgorithms.dynamicprogramming;
* cover the entire text ?-> matches single characters *-> match the sequence of * cover the entire text ?-> matches single characters *-> match the sequence of
* characters * characters
* *
* For calculation of Time and Space Complexity. Let N be length of src and M be * For calculation of Time and Space Complexity. Let N be length of src and M be length of pat
* length of pat
* *
* Memoization vs Tabulation : https://www.geeksforgeeks.org/tabulation-vs-memoization/
* Question Link : https://practice.geeksforgeeks.org/problems/wildcard-pattern-matching/1
*/ */
public final class RegexMatching { public final class RegexMatching {
private RegexMatching() { private RegexMatching() {
} }
// Method 1: Using Recursion /**
// Time Complexity=0(2^(N+M)) Space Complexity=Recursion Extra Space * Method 1: Determines if the given source string matches the given pattern using a recursive approach.
static boolean regexRecursion(String src, String pat) { * This method directly applies recursion to check if the source string matches the pattern, considering
* the wildcards '?' and '*'.
*
* Time Complexity: O(2^(N+M)), where N is the length of the source string and M is the length of the pattern.
* Space Complexity: O(N + M) due to the recursion stack.
*
* @param src The source string to be matched against the pattern.
* @param pat The pattern containing wildcards ('*' matches a sequence of characters, '?' matches a single character).
* @return {@code true} if the source string matches the pattern, {@code false} otherwise.
*/
public static boolean regexRecursion(String src, String pat) {
if (src.length() == 0 && pat.length() == 0) { if (src.length() == 0 && pat.length() == 0) {
return true; return true;
} }
@ -50,8 +61,19 @@ public final class RegexMatching {
return ans; return ans;
} }
// Method 2: Using Recursion and breaking string using virtual index /**
// Time Complexity=0(2^(N+M)) Space Complexity=Recursion Extra Space * Method 2: Determines if the given source string matches the given pattern using recursion.
* This method utilizes a virtual index for both the source string and the pattern to manage the recursion.
*
* Time Complexity: O(2^(N+M)) where N is the length of the source string and M is the length of the pattern.
* Space Complexity: O(N + M) due to the recursion stack.
*
* @param src The source string to be matched against the pattern.
* @param pat The pattern containing wildcards ('*' matches a sequence of characters, '?' matches a single character).
* @param svidx The current index in the source string.
* @param pvidx The current index in the pattern.
* @return {@code true} if the source string matches the pattern, {@code false} otherwise.
*/
static boolean regexRecursion(String src, String pat, int svidx, int pvidx) { static boolean regexRecursion(String src, String pat, int svidx, int pvidx) {
if (src.length() == svidx && pat.length() == pvidx) { if (src.length() == svidx && pat.length() == pvidx) {
return true; return true;
@ -83,9 +105,21 @@ public final class RegexMatching {
return ans; return ans;
} }
// Method 3: Top-Down DP(Memoization) /**
// Time Complexity=0(N*M) Space Complexity=0(N*M)+Recursion Extra Space * Method 3: Determines if the given source string matches the given pattern using top-down dynamic programming (memoization).
static boolean regexRecursion(String src, String pat, int svidx, int pvidx, int[][] strg) { * This method utilizes memoization to store intermediate results, reducing redundant computations and improving efficiency.
*
* Time Complexity: O(N * M), where N is the length of the source string and M is the length of the pattern.
* Space Complexity: O(N * M) for the memoization table, plus additional space for the recursion stack.
*
* @param src The source string to be matched against the pattern.
* @param pat The pattern containing wildcards ('*' matches a sequence of characters, '?' matches a single character).
* @param svidx The current index in the source string.
* @param pvidx The current index in the pattern.
* @param strg A 2D array used for memoization to store the results of subproblems.
* @return {@code true} if the source string matches the pattern, {@code false} otherwise.
*/
public static boolean regexRecursion(String src, String pat, int svidx, int pvidx, int[][] strg) {
if (src.length() == svidx && pat.length() == pvidx) { if (src.length() == svidx && pat.length() == pvidx) {
return true; return true;
} }
@ -120,8 +154,18 @@ public final class RegexMatching {
return ans; return ans;
} }
// Method 4: Bottom-Up DP(Tabulation) /**
// Time Complexity=0(N*M) Space Complexity=0(N*M) * Method 4: Determines if the given source string matches the given pattern using bottom-up dynamic programming (tabulation).
* This method builds a solution iteratively by filling out a table, where each cell represents whether a substring
* of the source string matches a substring of the pattern.
*
* Time Complexity: O(N * M), where N is the length of the source string and M is the length of the pattern.
* Space Complexity: O(N * M) for the table used in the tabulation process.
*
* @param src The source string to be matched against the pattern.
* @param pat The pattern containing wildcards ('*' matches a sequence of characters, '?' matches a single character).
* @return {@code true} if the source string matches the pattern, {@code false} otherwise.
*/
static boolean regexBU(String src, String pat) { static boolean regexBU(String src, String pat) {
boolean[][] strg = new boolean[src.length() + 1][pat.length() + 1]; boolean[][] strg = new boolean[src.length() + 1][pat.length() + 1];
strg[src.length()][pat.length()] = true; strg[src.length()][pat.length()] = true;
@ -153,15 +197,4 @@ public final class RegexMatching {
} }
return strg[0][0]; return strg[0][0];
} }
public static void main(String[] args) {
String src = "aa";
String pat = "*";
System.out.println("Method 1: " + regexRecursion(src, pat));
System.out.println("Method 2: " + regexRecursion(src, pat, 0, 0));
System.out.println("Method 3: " + regexRecursion(src, pat, 0, 0, new int[src.length()][pat.length()]));
System.out.println("Method 4: " + regexBU(src, pat));
} }
}
// Memoization vs Tabulation : https://www.geeksforgeeks.org/tabulation-vs-memoization/
// Question Link : https://practice.geeksforgeeks.org/problems/wildcard-pattern-matching/1

View File

@ -0,0 +1,43 @@
package com.thealgorithms.dynamicprogramming;
import static org.junit.jupiter.api.Assertions.assertEquals;
import java.util.stream.Stream;
import org.junit.jupiter.params.ParameterizedTest;
import org.junit.jupiter.params.provider.Arguments;
import org.junit.jupiter.params.provider.MethodSource;
public class RegexMatchingTest {
private record RegexTestCase(String s, String p, boolean expected) {
}
private static Stream<Arguments> provideTestCases() {
return Stream.of(Arguments.of(new RegexTestCase("aa", "*", true)), Arguments.of(new RegexTestCase("aa", "a*", true)), Arguments.of(new RegexTestCase("aa", "a", false)), Arguments.of(new RegexTestCase("cb", "?b", true)), Arguments.of(new RegexTestCase("cb", "?a", false)),
Arguments.of(new RegexTestCase("adceb", "*a*b", true)), Arguments.of(new RegexTestCase("acdcb", "a*c?b", false)), Arguments.of(new RegexTestCase("", "*", true)), Arguments.of(new RegexTestCase("", "", true)));
}
@ParameterizedTest
@MethodSource("provideTestCases")
void testRegexRecursionMethod1(RegexTestCase testCase) {
assertEquals(testCase.expected(), RegexMatching.regexRecursion(testCase.s(), testCase.p()));
}
@ParameterizedTest
@MethodSource("provideTestCases")
void testRegexRecursionMethod2(RegexTestCase testCase) {
assertEquals(testCase.expected(), RegexMatching.regexRecursion(testCase.s(), testCase.p(), 0, 0));
}
@ParameterizedTest
@MethodSource("provideTestCases")
void testRegexRecursionMethod3(RegexTestCase testCase) {
assertEquals(testCase.expected(), RegexMatching.regexRecursion(testCase.s(), testCase.p(), 0, 0, new int[testCase.s().length()][testCase.p().length()]));
}
@ParameterizedTest
@MethodSource("provideTestCases")
void testRegexBottomUp(RegexTestCase testCase) {
assertEquals(testCase.expected(), RegexMatching.regexBU(testCase.s(), testCase.p()));
}
}