Here is a standard Dynamic Programming solution for LeetCode 10 (Regular Expression Matching) in Java.ApproachWe use a 2D boolean array“dp[i][j]” which represents whether the first“i” characters of the string“s” match the first“j” characters of the pattern“p”.Key Rules:Base Case:“dp[0][0] true” (empty matches empty).Pattern ends with“: The” can match zero of the preceding element (“dp[i][j-2]”) OR one/more if the preceding character matches (“dp[i-1][j]”).Normal match /“‘.’”: If characters match or pattern is“‘.’”, carry over the previous state (“dp[i-1][j-1]”).Java Implementationclass Solution {public boolean isMatch(String s, String p) {int m s.length();int n p.length();// dp[i][j] true if first i chars of s match first j chars of p boolean[][] dp new boolean[m 1][n 1]; // 1. Base case: empty string matches empty pattern dp[0][0] true; // 2. Handle patterns like a*, a*b*, a*b*c* matching an empty string for (int j 2; j n; j) { if (p.charAt(j - 1) *) { dp[0][j] dp[0][j - 2]; } } // 3. Fill the DP table for (int i 1; i m; i) { for (int j 1; j n; j) { char currentCharS s.charAt(i - 1); char currentCharP p.charAt(j - 1); // Case A: Current characters match or pattern has . if (currentCharP currentCharS || currentCharP .) { dp[i][j] dp[i - 1][j - 1]; } // Case B: Current pattern character is * else if (currentCharP *) { // * matches zero of the preceding element dp[i][j] dp[i][j - 2]; // * matches one or more of the preceding element // Check if the character before * matches current string char char charBeforeStar p.charAt(j - 2); if (charBeforeStar currentCharS || charBeforeStar .) { dp[i][j] dp[i][j] || dp[i - 1][j]; } } // Case C: Characters dont match else { dp[i][j] false; } } } return dp[m][n]; }}Complexity AnalysisTime Complexity: O(m \times n) , where m is the length of“s” and nis the length ofp. We fill a table of size (m1) \times (n1)$.Space Complexity: O(m \times n) for the DP table. (This can be optimized to O(n) using a 1D array if needed.)Test Cases“s”“p” Output““aa””““a””“false”““aa””““a*””“true”““ab””“”.*“”“true”““aab””““cab””“true”““mississippi””““misisp*.””“false”Would you like me to explain a specific part of the logic in more detail, or provide the space-optimized (1D DP) version?