Solution: Regular Expression Matching
Statement▼
You are given an input string, s, and a pattern string, p. Your task is to implement regular expression matching between s and p, where the pattern may include the special characters '.' and '*':
'.'matches any single character.'*'matches zero or more occurrences of the preceding character.
The match must cover the entire input string, not just part of it.
Constraints:
1≤ s.length≤20 1≤ p.length≤20 scontains only lowercase English letters.pcontains only lowercase English letters,'.', and'*'.It is guaranteed that for each appearance of the character
'*', there will be a previous valid character to match.
Solution
One of the first ideas that comes to mind to solve the regular expression matching problem is to compare the string s and the pattern p character by character. At each step, we could check if the current characters match, and when we encounter special symbols like '.' or '*', we could try to handle them separately. However, this naive approach quickly becomes complicated because the '*' operator can represent zero or more repetitions of the previous character, leading to many possible interpretations of the pattern. A brute-force solution would involve exploring all possible combinations, resulting in an exponential time complexity.
Additionally, exploring all possible combinations requires performing numerous repeated calculations, i.e., checking the same parts of the string and pattern repeatedly in different recursive paths. For example, if the string is "aaa" and the pattern is "a*a", then while considering different ways the '*' could match one or more 'a's, the program repeatedly checks the same pair of positions, such as whether the last 'a' in the string matches the last 'a' in the pattern. These repeated computations represent overlapping subproblems, where the same question is solved multiple times unnecessarily, resulting in unnecessary computational effort.
Moreover, the problem has an optimal substructure, i.e., the final solution to the whole problem depends on solutions to its smaller subproblems. Whether the entire string and pattern match can be determined by knowing whether their suffixes match from certain points onward. For instance, whether "ab" matches "a*b" depends on whether the suffix 'b' matches 'b'. If the smaller part matches correctly, then the overall match can also succeed.
To handle these repeated calculations efficiently, we use recursion with memoization, which allows us to explore all valid ways to match the string s and pattern p while remembering already solved subproblems to avoid repeated work.
We use two pointers: one moving through the string s and the other through the pattern p. At each step, we ask: “Does the substring of s starting at the current index of the string pointer match the subpattern of p starting at the current index of the pattern pointer?”
The base case occurs when the pattern pointer reaches the end of p ...
Solution: Regular Expression Matching
Statement▼
You are given an input string, s, and a pattern string, p. Your task is to implement regular expression matching between s and p, where the pattern may include the special characters '.' and '*':
'.'matches any single character.'*'matches zero or more occurrences of the preceding character.
The match must cover the entire input string, not just part of it.
Constraints:
1≤ s.length≤20 1≤ p.length≤20 scontains only lowercase English letters.pcontains only lowercase English letters,'.', and'*'.It is guaranteed that for each appearance of the character
'*', there will be a previous valid character to match.
Solution
One of the first ideas that comes to mind to solve the regular expression matching problem is to compare the string s and the pattern p character by character. At each step, we could check if the current characters match, and when we encounter special symbols like '.' or '*', we could try to handle them separately. However, this naive approach quickly becomes complicated because the '*' operator can represent zero or more repetitions of the previous character, leading to many possible interpretations of the pattern. A brute-force solution would involve exploring all possible combinations, resulting in an exponential time complexity.
Additionally, exploring all possible combinations requires performing numerous repeated calculations, i.e., checking the same parts of the string and pattern repeatedly in different recursive paths. For example, if the string is "aaa" and the pattern is "a*a", then while considering different ways the '*' could match one or more 'a's, the program repeatedly checks the same pair of positions, such as whether the last 'a' in the string matches the last 'a' in the pattern. These repeated computations represent overlapping subproblems, where the same question is solved multiple times unnecessarily, resulting in unnecessary computational effort.
Moreover, the problem has an optimal substructure, i.e., the final solution to the whole problem depends on solutions to its smaller subproblems. Whether the entire string and pattern match can be determined by knowing whether their suffixes match from certain points onward. For instance, whether "ab" matches "a*b" depends on whether the suffix 'b' matches 'b'. If the smaller part matches correctly, then the overall match can also succeed.
To handle these repeated calculations efficiently, we use recursion with memoization, which allows us to explore all valid ways to match the string s and pattern p while remembering already solved subproblems to avoid repeated work.
We use two pointers: one moving through the string s and the other through the pattern p. At each step, we ask: “Does the substring of s starting at the current index of the string pointer match the subpattern of p starting at the current index of the pattern pointer?”
The base case occurs when the pattern pointer reaches the end of p ...