# Regular Expression Matching Problem

## Description

LeetCode Problem 10.

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:

``````1
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Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
``````

Example 2:

``````1
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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:

``````1
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Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
``````

Example 4:

``````1
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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:

``````1
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Input: s = "mississippi", p = "mis*is*p*."
Output: false
``````

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.

## Sample C++ Code

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class Solution {
public:
bool isMatch(string s, string p) {
int n = s.length();
int m = p.length();
vector<vector<int>> dp(n+1, vector<int>(m+1,0));
dp[0][0]=1;

for(int i=1; i<=m; i++){
if(p[i-1]=='*'){
dp[0][i]=dp[0][i-2];
}
}

for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){

if(s[i-1]==p[j-1] || p[j-1]=='.'){
dp[i][j]=dp[i-1][j-1];
}
else if(p[j-1]=='*'){
dp[i][j]=(dp[i][j-2] || (dp[i-1][j] && (p[j-2]=='.' || p[j-2]==s[i-1])));
}

}
}

return dp[n][m];
}
};
``````