In this post, we are going to solve the Regular Expression Matching Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
Problem
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 <= 201 <= p.length <= 30scontains only lowercase English letters.pcontains only lowercase English letters,'.', and'*'.- It is guaranteed for each appearance of the character
'*', there will be a previous valid character to match.
Now, let’s see the leetcode solution of Regular Expression Matching Leetcode Solution.
Regular Expression Matching Leetcode Solution in Python
class Solution:
def isMatch(self, s: str, p: str) -> bool:
s, p = ' '+ s, ' '+ p
lenS, lenP = len(s), len(p)
dp = [[0]*(lenP) for i in range(lenS)]
dp[0][0] = 1
for j in range(1, lenP):
if p[j] == '*':
dp[0][j] = dp[0][j-2]
for i in range(1, lenS):
for j in range(1, lenP):
if p[j] in {s[i], '.'}:
dp[i][j] = dp[i-1][j-1]
elif p[j] == "*":
dp[i][j] = dp[i][j-2] or int(dp[i-1][j] and p[j-1] in {s[i], '.'})
return bool(dp[-1][-1])Regular Expression Matching Leetcode Solution in CPP
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
dp[0][0] = true;
for (int i = 0; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (p[j - 1] == '*') {
dp[i][j] = dp[i][j - 2] || (i && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
} else {
dp[i][j] = i && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
}
}
}
return dp[m][n];
}
};Regular Expression Matching Leetcode Solution in Java
class Solution {
public boolean isMatch(String s, String p) {
if (p == null || p.length() == 0)
return (s == null || s.length() == 0);
boolean dp[][] = new boolean[s.length()+1][p.length()+1];
dp[0][0] = true;
for (int j=2; j<=p.length(); j++) {
dp[0][j] = p.charAt(j-1) == '*' && dp[0][j-2];
}
for (int j=1; j<=p.length(); j++) {
for (int i=1; i<=s.length(); i++) {
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] || ((s.charAt(i-1) == p.charAt(j-2) || p.charAt(j-2) == '.') && dp[i-1][j]);
}
}
return dp[s.length()][p.length()];
}
}Note: This problem Regular Expression Matching is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
