Leetcode solution 44: WildcardMatching
Leetcode solution 44: WildcardMatching
原文链接
https://blog.baozitraining.org/2019/04/leetcode-solution-44-wildcard-matching.html
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:
scould be empty and contains only lowercase lettersa-z.pcould be empty and contains only lowercase lettersa-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: falseProblem 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
- Initialize your lookup array, could be 2 dimensional or 1 dimensional using rolling arrays
- Initialize the init values in your array
- Get the math formula for a[i] = a[i-1] * m - n or whatever condition you need
- Code it up
You can find the video tutorial on this problem here
Solutions
Brute force recursion, TLE
public boolean isMatchTLE(String s, String p) { | |
|---|---|
if (s == null || p == null) { | |
return false; | |
} | |
return isMatchHelper(s, 0, p, 0); | |
} | |
// b*b*ab**ba*b**b***bba , this should trim some of the recursion time | |
public String removeDuplicateStars(String p) { | |
String newP = ""; | |
for (int i = 0; i < p.length(); i++) { | |
if (p.charAt(i) != '*') { | |
newP += p.charAt(i); | |
} else { | |
if (newP.length() == 0 || newP.charAt(newP.length() - 1) != '*') { | |
newP += "*"; | |
} | |
} | |
} | |
return newP; | |
} | |
// Even after all those optimizations, this would still TLE, n*2^m | |
private boolean isMatchHelper(String s, int i, String p, int j) { | |
if (j == p.length()) { | |
return i == s.length(); | |
} | |
p = this.removeDuplicateStars(p); | |
if (p.charAt(j) == '*') { | |
// s = "abcdef" | |
// p = "*bc*def" | |
if (isMatchHelper(s, i, p, j + 1)) { | |
return true; | |
} | |
if (i == s.length() - 1 && j == p.length() - 1) { | |
return true; | |
} | |
for (int k = i + 1; k < s.length(); k++) { | |
return isMatchHelper(s, k, p, j); | |
} | |
} else { | |
if (isCharMatch(s, i, p, j)) { | |
return isMatchHelper(s, i + 1, p, j + 1); | |
} | |
return false; | |
} | |
return false; | |
} | |
// compare if char is the same or it is a '?' | |
private boolean isCharMatch(String s, int i, String p, int j) { | |
if (i >= s.length() || j >= p.length()) { | |
return false; | |
} | |
return s.charAt(i) == p.charAt(j) || p.charAt(j) == '?'; | |
} |
view rawbruteforce.java hosted with ❤ by GitHub
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
// This is the DP solution, and falls perfectly using the DP template | |
|---|---|
// Also this is only boolean answer (not find all steps) || find the extreme values (min,max) | |
public boolean isMatch(String s, String p) { | |
if (s == null || p == null) { | |
return false; | |
} | |
int row = s.length(); | |
int col = p.length(); | |
boolean[][] lookup = new boolean[row + 1][col + 1]; | |
lookup[0][0] = true; | |
// initialize lookup init values, for the string (s) part, it all default to false | |
// since a char matches to empty is false anyway | |
for (int j = 1; j <= col; j++) { | |
if (p.charAt(j - 1) == '*' && lookup[0][j - 1]) { | |
lookup[0][j] = true; | |
} | |
} | |
for (int i = 1; i <= row; i++) { | |
for (int j = 1; j <= col; j++) { | |
if (p.charAt(j - 1) == '*') { | |
lookup[i][j] = lookup[i][j - 1] || lookup[i - 1][j]; | |
} else { | |
if (lookup[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?')) { | |
lookup[i][j] = true; | |
} | |
} | |
} | |
} | |
return lookup[row][col]; | |
} |
view rawwildcardMatching.java hosted with ❤ by GitHub
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