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Regular Expression Matching Leetcode Solution

Regular Expression Matching





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 <= 20
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.

Solution : 

class Solution {
public:
    bool isMatch(string s, string p) {

        vector<vector<bool>>dp(p.length() +1 , vector<bool>(s.length()+1 , false));
        for(int i =0 ; i < p.length() +1 ; i++)
        {
            for(int j =0 ; j < s.length() +1 ; j++)
            {
                if(i ==0 && j ==0)
                {
                    dp[i][j] = true;
                }
                else
                if(i ==0)
                {
                    dp[i][j] = false;
                }
                else
                if(j ==0)
                {
                    char pc = p[i-1];
                 
                    if(pc == '*')
                    {
                        dp[i][j] = dp[i-2][j];
                    }

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

                }
            }
         
        }
 

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

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