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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).
Test Cases
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 <= 201 <= p.length <= 20scontains only lowercase English letters.pcontains 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 boolean isMatch(String text, String pattern) {
boolean[][] dp = new boolean[text.length() + 1][pattern.length() + 1];
dp[text.length()][pattern.length()] = true;
for (int i = text.length(); i >= 0; i--){
for (int j = pattern.length() - 1; j >= 0; j--){
boolean first_match = (i < text.length() &&
(pattern.charAt(j) == text.charAt(i) ||
pattern.charAt(j) == '.'));
if (j + 1 < pattern.length() && pattern.charAt(j+1) == '*'){
dp[i][j] = dp[i][j+2] || first_match && dp[i+1][j];
} else {
dp[i][j] = first_match && dp[i+1][j+1];
}
}
}
return dp[0][0];
}
}class Solution:
def isMatch(self, s: str, p: str) -> bool:
lp, ls = len(p), len(s)
dp = [[False] * (lp+1) for _ in range(ls + 1)]
dp[-1][-1] = True
for i in range(ls, -1, -1):
for j in range(lp - 1, -1, -1):
match = i < ls and p[j] in {s[i], '.'}
if j + 1 < lp and p[j+1] == '*':
dp[i][j] = dp[i][j+2] or match and dp[i+1][j]
else:
dp[i][j] = match and dp[i+1][j+1]
return dp[0][0]