# Regular Expression Matching

#1

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

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

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

Note:

• `s` could be empty and contains only lowercase letters `a-z`.
• `p` could be empty and contains only lowercase letters `a-z`, and characters like `.` or `*`.

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 precedeng 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 (.)".
```

Example 4:

```Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
```

Example 5:

```Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
```

#2

The code below uses dynamic programming to find a certain pattern matches a string.

Below are the rules that drives the DP logic.

``````1, If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == '*':
here are two sub conditions:
1   if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty
2   if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
dp[i][j] = dp[i-1][j]    //in this case, a* counts as multiple a
or dp[i][j] = dp[i][j-1]   // in this case, a* counts as single a
or dp[i][j] = dp[i][j-2]   // in this case, a* counts as empty
``````
``````class Solution {
public boolean isMatch(String s, String p) {
if(s==null || p==null) return true;
int m=s.length(), n=p.length();
boolean[][] dp = new boolean[m+1][n+1];
dp[0][0] = true;
// sets all indexes which has a '*' to 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<m;i++){
for(int j=0;j<n;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(s.charAt(i) != p.charAt(j-1) && 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[m][n];
}
}
``````