# Wildcard Matching

Wildcard Matching

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

**Dynamic Programming**

Similar to Regular Expression Matching, the difference is how the

`'*'`

, `'?'`

are interpreted, also it uses `'?'`

instead of `'.'`

Yet the fundamentals are the same, just the initialization and state transfer function needs to be modified.

Although it might be easier to think of DP solution if you've met Regular Expression Matching problem before, there's a more straightforward approach that's more intuitive.

DP - O(mn) space, O(mn) time - (33ms 76% AC)

class Solution {

public boolean isMatch(String s, String p) {

char[] str = s.toCharArray();

char[] pattern = p.toCharArray();

int m = str.length;

int n = pattern.length;

boolean dp[][] = new boolean[m + 1][n + 1];

dp[0][0] = true;

for (int i = 1; i <= n; i++) {

if (pattern[i - 1] == '*') {

dp[0][i] = dp[0][i - 1];

}

}

for (int i = 1; i <= m; i++) {

for (int j = 1; j <= n; j++) {

if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '?') {

dp[i][j] = dp[i - 1][j - 1];

}

if (pattern[j - 1] == '*') {

dp[i][j] = dp[i - 1][j] || dp[i][j - 1] || dp[i - 1][j - 1];

}

}

}

return dp[m][n];

}

}

Multi-Pointers - O(1) space, O(mn) time

boolean comparison(String str, String pattern) {

int s = 0, p = 0, match = 0, starIdx = -1;

while (s < str.length()){

// advancing both pointers

if (p < pattern.length() && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){

s++;

p++;

}

// * found, only advancing pattern pointer

else if (p < pattern.length() && pattern.charAt(p) == '*'){

starIdx = p;

match = s;

p++;

}

// last pattern pointer was *, advancing string pointer

else if (starIdx != -1){

p = starIdx + 1;

match++;

s = match;

}

//current pattern pointer is not star, last patter pointer was not *

//characters do not match

else return false;

}

//check for remaining characters in pattern

while (p < pattern.length() && pattern.charAt(p) == '*')

p++;

return p == pattern.length();

}

Last modified 2yr ago