Leetcode solution 44: Wildcard Matching

Problem Statement 

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like ? or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

Problem link

Video Tutorial

You can find the detailed video tutorial here

Thought Process

It seems an easy problem first using recursion and brute forcefully solving the problem, but the OJ will quickly throw a TLE (Time Limit Exceeded) error since it is exponential time. Well, if you were like me, tried really hard to optimize my brute force solution, either using some heuristics or pruning on the recursion, however, still no luck :(

DP (Dynamic Programming) is normally a great way when trying to answer below two types of questions, and string related questions are a majority use case of DP

• Only Yes/No, True/False question without returning detailed step by step results 
• Extreme values, e.g., Max, Min, Best optimal value

 I have posted several videos on how to solve DP problems using a template, essentially

1. Initialize your lookup array, could be 2 dimensional or 1 dimensional using rolling arrays
2. Initialize the init values in your array
3. Get the math formula for a[i] = a[i-1] * m - n or whatever condition you need
4. Code it up 

You can find the video tutorial on this problem here

Solutions

Brute force recursion, TLE

Time Complexity: assuming S length is M, P length is N, this is O(M*2^N), exponential basically
Space Complexity: No extra space needed other than the additional recursion function stack

 

DP solution

Time Complexity: assuming S length is M, P length is N, this is O(M*N) because all we need to do is build up that lookup matrix
Space Complexity: assuming S length is M, P length is N, this is O(M*N), again, that lookup matrix

References

• A good DP video tutorial
• Grandyang very funny writings in Chinese
• Yu's garden, a not very well explained but very clever(hacky) solution