A few days ago I bumped into a question on LeetCode where I saw a bunch of people scratching their heads. It was question number 10 titled Regular Expression Matching.
A few days ago I bumped into a question on LeetCode where I saw a bunch of people scratching their heads. It was question number 10 titled Regular Expression Matching with the difficulty label
Hard. The solution to the exercise itself was not intuitive, requiring effective use of previously solved subproblems, but that wasn’t what everyone in the comments were interested in. After all the solution was available for everyone to see.
The real intrigue was the calculated but unexplained time complexity for one of the inefficient solutions which only used recursion, minus the dynamic programming. After all, dynamic programming makes time and space complexity fairly easy to compute, as long as the solution is correct, it cannot exceed the complexity of all possible subproblems. Before I can get ahead of myself though here’s the outline of the original problem which can also be found here https://leetcode.com/problems/regular-expression-matching/
If at some point I lose you during the explanation please leave a comment so I can make a revision. I’m not used to writing these.
Given an input string (
s) and a pattern (
p), implement regular expression matching with support for
'.' Matches any single character. '*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
scould be empty and contains only lowercase letters
pcould be empty and contains only lowercase letters
a-z, and characters like
Input: s = "abcda" p = "ab.d" Output: false Explanation: The . matches the c but the pattern ends before matching to the entire text.
Input: s = "aab" p = "c*a*b" Output: true Explanation: c can be repeated 0 times, a can be repeated 2 times. Therefore, it matches "aab".
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