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# Regular Expression Matching LeetCode Solution

## Problem – 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 <= 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.

### Regular Expression Matching LeetCode Solution in Java

public boolean isMatch(String s, String p) {

if (s == null || p == null) {
return false;
}
boolean[][] dp = new boolean[s.length()+1][p.length()+1];
dp[0][0] = 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 < s.length(); i++) {
for (int j = 0; j < p.length(); 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 (p.charAt(j-1) != s.charAt(i) && 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[s.length()][p.length()];
}

### Regular Expression Matching LeetCode Solution in C++

class Solution {
public:
bool isMatch(string s, string p) {
if (p.empty())    return s.empty();

if ('*' == p[1])
// x* matches empty string or at least one character: x* -> xx*
// *s is to ensure s is non-empty
return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
else
return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
}
};

class Solution {
public:
bool isMatch(string s, string p) {
/**
* f[i][j]: if s[0..i-1] matches p[0..j-1]
* if p[j - 1] != '*'
*      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
* if p[j - 1] == '*', denote p[j - 2] with x
*      f[i][j] is true iff any of the following is true
*      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
*      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
* '.' matches any single character
*/
int m = s.size(), n = p.size();
vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));

f[0][0] = true;
for (int i = 1; i <= m; i++)
f[i][0] = false;
// p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
for (int j = 1; j <= n; j++)
f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];

for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
if (p[j - 1] != '*')
f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
else
// p[0] cannot be '*' so no need to check "j > 1" here
f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];

return f[m][n];
}
};

### Regular Expression Matching LeetCode Solution in Python

import unittest

class Solution(object):
def isMatch(self, s, p):
# The DP table and the string s and p use the same indexes i and j, but
# table[i][j] means the match status between p[:i] and s[:j], i.e.
# table[0][0] means the match status of two empty strings, and
# table[1][1] means the match status of p[0] and s[0]. Therefore, when
# refering to the i-th and the j-th characters of p and s for updating
# table[i][j], we use p[i - 1] and s[j - 1].

# Initialize the table with False. The first row is satisfied.
table = [[False] * (len(s) + 1) for _ in range(len(p) + 1)]

# Update the corner case of matching two empty strings.
table[0][0] = True

# Update the corner case of when s is an empty string but p is not.
# Since each '*' can eliminate the charter before it, the table is
# vertically updated by the one before previous. [test_symbol_0]
for i in range(2, len(p) + 1):
table[i][0] = table[i - 2][0] and p[i - 1] == '*'

for i in range(1, len(p) + 1):
for j in range(1, len(s) + 1):
if p[i - 1] != "*":
# Update the table by referring the diagonal element.
table[i][j] = table[i - 1][j - 1] and \
(p[i - 1] == s[j - 1] or p[i - 1] == '.')
else:
# Eliminations (referring to the vertical element)
# Either refer to the one before previous or the previous.
# I.e. * eliminate the previous or count the previous.
# [test_symbol_1]
table[i][j] = table[i - 2][j] or table[i - 1][j]

# Propagations (referring to the horizontal element)
# If p's previous one is equal to the current s, with
# helps of *, the status can be propagated from the left.
# [test_symbol_2]
if p[i - 2] == s[j - 1] or p[i - 2] == '.':
table[i][j] |= table[i][j - 1]

return table[-1][-1]

class TestSolution(unittest.TestCase):
def test_none_0(self):
s = ""
p = ""
self.assertTrue(Solution().isMatch(s, p))

def test_none_1(self):
s = ""
p = "a"
self.assertFalse(Solution().isMatch(s, p))

def test_no_symbol_equal(self):
s = "abcd"
p = "abcd"
self.assertTrue(Solution().isMatch(s, p))

def test_no_symbol_not_equal_0(self):
s = "abcd"
p = "efgh"
self.assertFalse(Solution().isMatch(s, p))

def test_no_symbol_not_equal_1(self):
s = "ab"
p = "abb"
self.assertFalse(Solution().isMatch(s, p))

def test_symbol_0(self):
s = ""
p = "a*"
self.assertTrue(Solution().isMatch(s, p))

def test_symbol_1(self):
s = "a"
p = "ab*"
self.assertTrue(Solution().isMatch(s, p))

def test_symbol_2(self):
# E.g.
#   s a b b
# p 1 0 0 0
# a 0 1 0 0
# b 0 0 1 0
# * 0 1 1 1
s = "abb"
p = "ab*"
self.assertTrue(Solution().isMatch(s, p))

if __name__ == "__main__":
unittest.main()
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