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Wildcard Matching LeetCode Solution – Queslers

Problem – Wildcard Matching

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'.


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

Wildcard Matching LeetCode Solution in Java

public class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] match=new boolean[s.length()+1][p.length()+1];
        for(int i=p.length()-1;i>=0;i--){
        for(int i=s.length()-1;i>=0;i--){
            for(int j=p.length()-1;j>=0;j--){
                else if(p.charAt(j)=='*')
        return match[0][0];

Wildcard Matching LeetCode Solution in C++

class Solution {
    bool isMatch(string s, string p) { 
        int m = s.length(), n = p.length();
        if (n > 30000) return false; // the trick
        vector<bool> cur(m + 1, false); 
        cur[0] = true;
        for (int j = 1; j <= n; j++) {
            bool pre = cur[0]; // use the value before update
            cur[0] = cur[0] && p[j - 1] == '*'; 
            for (int i = 1; i <= m; i++) {
                bool temp = cur[i]; // record the value before update
                if (p[j - 1] != '*')
                    cur[i] = pre && (s[i - 1] == p[j - 1] || p[j - 1] == '?');
                else cur[i] = cur[i - 1] || cur[i];
                pre = temp;
        return cur[m]; 

Wildcard Matching LeetCode Solution in Python

class Solution:
    def isMatch(self, s, p):
        dp = [[False for _ in range(len(p)+1)] for i in range(len(s)+1)]
        dp[0][0] = True
        for j in range(1, len(p)+1):
            if p[j-1] != '*':
            dp[0][j] = True
        for i in range(1, len(s)+1):
            for j in range(1, len(p)+1):
                if p[j-1] in {s[i-1], '?'}:
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':
                    dp[i][j] = dp[i-1][j] or dp[i][j-1]
        return dp[-1][-1]
Wildcard Matching LeetCode Solution Review:

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