# Regular Expression Matching

This page explains Java solution to problem `Regular Expression Matching` using `Dynamic Programming`.

## Problem Statement

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.

Example 1:

Input: s = "aa", p = "a"
Output: false

Example 2:

Input: s = "aab", p = "c*a*b"
Output: true

Example 3:

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

## Solution

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``````
package com.vc.hard;

class RegularExpressionMatching {
public boolean isMatch(String s, String p) {
if(s == null && p == null) return true;
if(s == null || p == null) return false;

int m = s.length();
int n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
for(int si = 0; si <= m; si++) {
for(int pi = 0; pi <= n; pi++) {
if(si == 0 && pi == 0) {
//if length of input string si & input pattern pi is 0
dp[si][pi] = true;
}
else if(pi == 0) {
//if length of input pattern pi is 0
dp[si][pi] = false;
}
else if(si == 0) {
//if length of input string si is 0
//here pi will be > 0, so we have to handle special cases * and .
if(p.charAt(pi - 1) == '*') {
dp[si][pi] = dp[si][pi - 2];
}
else if(p.charAt(pi - 1) == '.') {
dp[si][pi] = false;
}
else {
dp[si][pi] = false;
}
}
else {
//if length of input string as well as input pattern is > 0
//here pi will be > 0, so we have to handle special cases * and .
if(p.charAt(pi - 1) == '*')  {
dp[si][pi] = dp[si][pi - 2];
if(p.charAt(pi - 2) == '.' || p.charAt(pi - 2) == s.charAt(si - 1)) {
dp[si][pi] |= dp[si - 1][pi];
}
}
else if(p.charAt(pi - 1) == '.') {
dp[si][pi] = dp[si - 1][pi - 1];
}
else if(p.charAt(pi - 1) == s.charAt(si - 1)) {
dp[si][pi] = dp[si - 1][pi - 1];
}
else {
dp[si][pi] = false;
}
}
}
}
return dp[m][n];
}
}
``````

O(m * n)

O(m * n)