Wildcard Matching


Problem Statement :


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 '*'.



Solution :



title-img


                            Solution in C :

bool isMatch(char * s, char * p){
    int m = strlen(s), n = strlen(p);
    int i = 0, j = 0;
    int star_idx = -1, s_idx = -1;
    while (i < m) {
        if (j < n && (s[i] == p[j] || p[j] == '?')) {
            i++;
            j++;
        } else if (j < n && p[j] == '*') {
            star_idx = j;
            s_idx = i;
            j++;
        } else if (star_idx != -1) {
            j = star_idx + 1;
            i = s_idx + 1;
            s_idx++;
        } else {
            return false;
        }
    }
    while (j < n && p[j] == '*') {
        j++;
    }
    return j == n;
}
                        


                        Solution in C++ :

class Solution {
public:
    
    bool isMatch(string s, string p) {
        int n = s.size();
        int m = p.size();
        vector<vector<bool>> dp(n+1, vector<bool>(m+1, false));
        dp[0][0] = true;
        if(p[0] == '*') dp[0][1] = true;
        for(int j=2; j<=m; j++){
            if(p[j-1] == '*') dp[0][j] = dp[0][j-1];
        }

        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; 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[n][m];
    }
};
                    


                        Solution in Java :

// TC -> O(|s|*|p|)
// SC -> O(|s|*|p|)

class Solution {
    public boolean isMatch(String s, String p) {
        return solve(s,p,0,0,new Boolean[s.length()][p.length()]);
    }

    private boolean solve(String s,String t,int i,int j,Boolean[][] dp){
        if(i==s.length()){
            return hasNoLCLeft(t,j);
        }
        if(j==t.length()){
            return i==s.length();
        }
        if(dp[i][j]!=null){
            return dp[i][j];
        }
        if(Character.isLowerCase(t.charAt(j)) && s.charAt(i)!=t.charAt(j)){
            return dp[i][j] = false;
        }
        if(t.charAt(j)=='*'){
            if(solve(s,t,i+1,j,dp)){
                return true;
            }
            if(solve(s,t,i,j+1,dp)){
                return true;
            }
        }
        return dp[i][j] = solve(s,t,i+1,j+1,dp)? true: false;
    }


    private boolean hasNoLCLeft(String t,int j){
        for(int i=j;i<t.length();i++){
            if(t.charAt(i)!='*'){
                return false;
            }
        }
        return true;
    }
}
                    


                        Solution in Python : 
                            
# DP
    def isMatch(self, s: str, p: str) -> bool:
        dp = [[False]*(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] = dp[0][j-1]
        
        for i in range(1,len(s)+1):
            for j in range(1,len(p)+1):
                dp[i][j] = (p[j-1] in [s[i-1],'?','*'] and dp[i-1][j-1]) or (p[j-1] == '*' and (dp[i][j-1] or dp[i-1][j]))
  
        return dp[len(s)][len(p)]
                    


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