Wildcard Pattern

#1

Sir,
Can you please tell me what is error in my code

#2

Pls share your code by saving on ide so that I could check it…

#3

here is the code https://ide.codingblocks.com/s/293360

#4

Pls follow the approach as:

bool wildcard(string text,string pattern){
int r = text.size();
int c = pattern.size();

bool dp[r+1][c+1];          //dp[i][j] denotes if the text till i and pattern till j match

for(int i=0;i<=r;i++){
    dp[i][0] = dp[0][i] = false  ;
}

dp[0][0] = true ; //Base case - Strings of 0 length always match

//If first character of pattern is * then mark (0,1) as true since * can be mapped with empty sequence as well
if(pattern[0] == '*'){
    dp[0][1] = true ;
}
for(int i=0;pattern[i]=='*';i++){
    dp[i][1] = true ; 
}

for(int i=1;i<=r;i++){
    for(int j=1;j<=c;j++){
        if(text[i-1] == pattern[j-1]){
            //If current character of each string matches then mark this as however its previous state was
            dp[i][j] = dp[i-1][j-1];
        }
        else if(pattern[j-1] == '?'){
            //Since ? can be mapped with all characters , do the same for this as well
            dp[i][j] = dp[i-1][j-1];
        }
        else if(pattern[j-1] == '*'){
            //Since * can be mapped with all strings , mark current state as OR of both previous states
            dp[i][j] = dp[i-1][j] || dp[i][j-1] ; 
        }
        else {
            //Since no matching took place , mark this cell as false
            dp[i][j] = false ;
        }
    }
}
return dp[r][c];

}

#5

Sir i submitted solution but could not understand the error in my code so i downloaded the test cases.