**Regular Expression Matching - Amazon Top Interview Questions**

### Problem Statement :

Implement regular expression matching with the following special characters: . (period) which matches any single character * (asterisk) which matches zero or more of the preceding element That is, implement a function that takes in a valid regular expression pattern and a string s and returns whether or not the string matches the regular expression. Note: The input pattern is guaranteed not to have consecutive asterisks. Constraints n ≤ 100 where n is the length of pattern m ≤ 1,000 where m is the length of s Example 1 Input pattern = "ra." s = "ray" Output True Explanation We have ra and then a single character Example 2 Input pattern = "a" s = "aa" Output False Example 3 Input pattern = "a*" s = "aa" Output True Explanation We have 0 or more as. Example 4 Input pattern = ".*" s = "abc" Output True Explanation We have 0 or more of any character

### Solution :

` ````
Solution in C++ :
int dp[101][1001];
bool matching(string &pattern, string &s, int i, int j) {
if (i == pattern.length() && j == s.length()) return true;
if (i == pattern.length()) return false;
if (dp[i][j] != -1) return dp[i][j];
if (pattern[i] == s[j] || pattern[i] == '.') {
if (pattern[i + 1] == '*' && j + 1 < s.length())
return dp[i][j] = matching(pattern, s, i + 2, j) || matching(pattern, s, i, j + 1);
else if (pattern[i + 1] == '*' && j + 1 >= s.length())
return dp[i][j] = matching(pattern, s, i + 2, j + 1) || matching(pattern, s, i + 2, j);
else
return dp[i][j] = matching(pattern, s, i + 1, j + 1);
} else {
if (pattern[i + 1] == '*')
return dp[i][j] = matching(pattern, s, i + 2, j);
else
return dp[i][j] = false;
}
}
bool solve(string pattern, string s) {
memset(dp, -1, sizeof(dp));
return matching(pattern, s, 0, 0);
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(final String pattern, final String text) {
if (pattern == null || pattern.length() == 0) {
return text == null || text.length() == 0;
}
if (text == null || text.length() == 0) {
return false;
}
Boolean[][] match = new Boolean[text.length() + 1][pattern.length() + 1];
return isMatch(text, pattern, 0, 0, match);
}
private boolean isMatch(final String text, final String pattern, final int tL, final int pL,
final Boolean[][] match) {
if (pL == pattern.length()) {
return tL == text.length();
}
if (match[tL][pL] != null) {
return match[tL][pL];
}
boolean firstMatch = (tL < text.length()
&& (text.charAt(tL) == pattern.charAt(pL) || pattern.charAt(pL) == '.'));
boolean result = false;
if (pL + 1 < pattern.length() && pattern.charAt(pL + 1) == '*') {
result = isMatch(text, pattern, tL, pL + 2, match)
|| (firstMatch && isMatch(text, pattern, tL + 1, pL, match));
} else {
result = firstMatch && isMatch(text, pattern, tL + 1, pL + 1, match);
}
match[tL][pL] = result;
return match[tL][pL];
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, pattern, s):
m = len(pattern)
n = len(s)
dp = [[False for _ in range(m + 1)] for _ in range(n + 1)]
dp[n][m] = True
for i in range(m - 1, -1, -1):
if i + 1 < m and pattern[i + 1] == "*":
dp[n][i] = dp[n][i + 2]
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
if i + 1 < m and pattern[i + 1] == "*":
dp[j][i] = dp[j][i + 2]
if s[j] == pattern[i] or pattern[i] == ".":
dp[j][i] = dp[j][i + 2] | dp[j + 1][i]
else:
if pattern[i] == s[j] or pattern[i] == ".":
dp[j][i] = dp[j + 1][i + 1]
# print(i, j, dp[j][i])
return dp[0][0]
```

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