Add hard problems

Author: das-jishu

Leetcode/hard/regular-expression-matching.py

@@ -0,0 +1,78 @@
+"""
+# 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 (.)".
+Example 4:
+
+Input: s = "aab", p = "c*a*b"
+Output: true
+Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
+Example 5:
+
+Input: s = "mississippi", p = "mis*is*p*."
+Output: false
+ 
+
+Constraints:
+
+0 <= s.length <= 20
+0 <= 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.
+"""
+
+class Solution:
+    def isMatch(self, s: str, p: str) -> bool:
+        
+        table = [[None for _ in range(len(p) + 1)] for _ in range(len(s) + 1)]
+        table[0][0] = True
+        
+        
+        for row in range(1, len(s) + 1):
+            table[row][0] = False
+            
+        for col in range(1, len(p) + 1):
+            if p[col - 1] == "*":
+                table[0][col] = table[0][col - 2]
+            else:
+                table[0][col] = False
+            
+        for row in range(1, len(s) + 1):
+            for col in range(1, len(p) + 1):
+                if p[col - 1] == ".":
+                    table[row][col] = table[row - 1][col - 1]
+                elif p[col - 1] != "*":
+                    table[row][col] = table[row - 1][col - 1] and s[row - 1] == p[col - 1]
+                else:
+                    if p[col - 2] == ".":
+                        table[row][col] = table[row][col - 1] or table[row][col - 2] or table[row - 1][col - 1] or table[row - 1][col]
+                    else:
+                        table[row][col] = table[row][col - 1] or table[row][col - 2] or (table[row - 1][col - 1] and p[col - 2] == s[row - 1])
+                    
+        for row in range(len(s) + 1):
+            print(table[row])
+                    
+        return table[-1][-1]