# Regular Expression Matching – Leetcode Solution

In this post, we are going to solve the 10. Regular Expression Matching problem of Leetcode. This problem 10. Regular Expression Matching is a Leetcode hard level problem. Let’s see code, 10. Regular Expression Matching.

Contents

## Problem

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

• `'.'` Matches any single character.â€‹â€‹â€‹â€‹
• `'*'` 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.

Now, let’s see the code of 10. Regular Expression Matching – Leetcode Solution.

# Regular Expression Matching – Leetcode Solution

### 10. Regular Expression Matching – 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()];
}
}```

### 10. Regular Expression Matching – Solution in C++

```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];
}
};```

### 10. Regular Expression Matching– 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])
```

Note: This problem 10. Regular Expression Matching is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.