# 44. Wildcard Matching

## Description

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

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

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

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'.


Constraints:

• 0 <= s.length, p.length <= 2000
• s contains only lowercase English letters.
• p contains only lowercase English letters, '?' or '*'.

## Solutions

Solution 1: Dynamic Programming

We define the state $dp[i][j]$ to represent whether the first $i$ characters of $s$ match the first $j$ characters of $p$.

The state transition equation is as follows:

$dp[i][j]= \begin{cases} dp[i-1][j-1] & \text{if } s[i-1]=p[j-1] \text{ or } p[j-1]=\text{?} \\ dp[i-1][j-1] \lor dp[i-1][j] \lor dp[i][j-1] & \text{if } p[j-1]=\text{*} \\ \text{false} & \text{otherwise} \end{cases}$

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of $s$ and $p$ respectively.

• class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
dp[0][j] = dp[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p.charAt(j - 1) == '*') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}
}

• class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
vector<vector<bool>> dp(m + 1, vector<bool>(n + 1));
dp[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
dp[0][j] = dp[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s[i - 1] == p[j - 1] || p[j - 1] == '?') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p[j - 1] == '*') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}
};

• class Solution:
def isMatch(self, s: str, p: str) -> bool:
m, n = len(s), len(p)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True

# if p starting with "*", then all true for dp[0][i]
# or else, all false for dp[0][i]
for j in range(1, n + 1):
if p[j - 1] == '*':
dp[0][j] = dp[0][j - 1]

for i in range(1, m + 1):
for j in range(1, n + 1):
if s[i - 1] == p[j - 1] or p[j - 1] == '?':
dp[i][j] = dp[i - 1][j - 1]
# dp[i - 1][j], where j is always '*', all the way back to i-1==0
elif p[j - 1] == '*':
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
return dp[m][n]


• func isMatch(s string, p string) bool {
m, n := len(s), len(p)
dp := make([][]bool, m+1)
for i := range dp {
dp[i] = make([]bool, n+1)
}
dp[0][0] = true
for j := 1; j <= n; j++ {
if p[j-1] == '*' {
dp[0][j] = dp[0][j-1]
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if s[i-1] == p[j-1] || p[j-1] == '?' {
dp[i][j] = dp[i-1][j-1]
} else if p[j-1] == '*' {
dp[i][j] = dp[i-1][j] || dp[i][j-1]
}
}
}
return dp[m][n]
}

• using System.Linq;

public class Solution {
public bool IsMatch(string s, string p) {
if (p.Count(ch => ch != '*') > s.Length)
{
return false;
}

bool[,] f = new bool[s.Length + 1, p.Length + 1];
bool[] d = new bool[s.Length + 1]; // d[i] means f[0, j] || f[1, j] || ... || f[i, j]
for (var j = 0; j <= p.Length; ++j)
{
d[0] = j == 0 ? true : d[0] && p[j - 1] == '*';
for (var i = 0; i <= s.Length; ++i)
{
if (j == 0)
{
f[i, j] = i == 0;
continue;
}

if (p[j - 1] == '*')
{
if (i > 0)
{
d[i] = f[i, j - 1] || d[i - 1];
}
f[i, j] = d[i];
}
else if (p[j - 1] == '?')
{
f[i, j] = i > 0 && f[i - 1, j - 1];
}
else
{
f[i, j] = i > 0 && f[i - 1, j - 1] && s[i - 1] == p[j - 1];
}
}
}
return f[s.Length, p.Length];
}
}

• function isMatch(s: string, p: string): boolean {
const m = s.length;
const n = p.length;
const f: number[][] = Array.from({ length: m + 1 }, () =>
Array.from({ length: n + 1 }, () => -1),
);
const dfs = (i: number, j: number): boolean => {
if (i >= m) {
return j >= n || (p[j] === '*' && dfs(i, j + 1));
}
if (j >= n) {
return false;
}
if (f[i][j] !== -1) {
return f[i][j] === 1;
}
if (p[j] === '*') {
f[i][j] = dfs(i + 1, j) || dfs(i, j + 1) ? 1 : 0;
} else {
f[i][j] = (p[j] === '?' || s[i] === p[j]) && dfs(i + 1, j + 1) ? 1 : 0;
}
return f[i][j] === 1;
};
return dfs(0, 0);
}


• class Solution {
/**
* @param string $s * @param string$p
* @return boolean
*/

function isMatch($s,$p) {
$lengthS = strlen($s);
$lengthP = strlen($p);
$dp = array(); for ($i = 0; $i <=$lengthS; $i++) {$dp[$i] = array_fill(0,$lengthP + 1, false);
}
$dp[0][0] = true; for ($i = 1; $i <=$lengthP; $i++) { if ($p[$i - 1] == '*') {$dp[0][$i] =$dp[0][$i - 1]; } } for ($i = 1; $i <=$lengthS; $i++) { for ($j = 1; $j <=$lengthP; $j++) { if ($p[$j - 1] == '?' ||$s[$i - 1] ==$p[$j - 1]) {$dp[$i][$j] = $dp[$i - 1][$j - 1]; } else if ($p[$j - 1] == '*') {$dp[$i][$j] = $dp[$i][$j - 1] ||$dp[$i - 1][$j];
}
}
}
return $dp[$lengthS][\$lengthP];
}
}