Leetcode 10 Regular Expression Matching Solution in Java | Hindi Coding Community

 Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

 

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

 

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

Java Code :

public boolean isMatch(String s, String p) {

if (s == null || p == null) {

return false;

}

boolean[][] dp = new boolean[s.length()+1][p.length()+1];

dp[0][0] = true;

for (int i = 0; i < p.length(); i++) {

if (p.charAt(i) == '*' && dp[0][i-1]) {

dp[0][i+1] = true;

}

}

for (int i = 0 ; i < s.length(); i++) {

for (int j = 0; j < p.length(); j++) {

if (p.charAt(j) == '.') {

dp[i+1][j+1] = dp[i][j];

}

if (p.charAt(j) == s.charAt(i)) {

dp[i+1][j+1] = dp[i][j];

}

if (p.charAt(j) == '*') {

if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {

dp[i+1][j+1] = dp[i+1][j-1];

} else {

dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);

}

}

}

}

return dp[s.length()][p.length()];

}