CareerCup

This is from a programming contest/website.

haha, I choose Perl to implement this:

$exp = q(^ab..*abc$);
$test = q(abyxxxxabc);

print "$exp, $test, " . evaluate($exp, $test)."\n";

sub evaluate{
	my $Exp = shift;
	my $testCase = shift;

	return "false" if ($testCase !~ /^[a-z]+[a-z]$/);	# contain only character in [a-z]?

	if ($testCase =~ $Exp){
		return "true";
	}
	else{
		return "false";
	}
};

^ and $ can be neglected removed as string without same also mean the same thing. Here is a dyanamic approach to the problem

bool isMatch(const char *s, const char *p) {
        int n=strlen(s), m=strlen(p), i, j, chars=0;
    
        //check if pattern have less that equal to character then string
        for(i=0; p[i]!='\0'; ++i) if(p[i]!='*' && n<++chars) return false;
        
        vector<bool> dp(n+2,false);
        // dp[n] is true fo first time, 
        // i.e end of string charater is true to enable correct match for last pattern character
        // At any time dp[i] == 1 mean we have match from i to n index
        for(i=m-1, dp[n]=true; i>=0; --i){
            if(p[i]=='*'){
                while(i>0 && p[i]==p[i-1]) --i; //skip multiple *
                for(j=n; j>=0 && !dp[j]; --j);
                for(; j>=0; --j) dp[j]=true;
            }else{
                for(j=0; j<n+1; ++j)
                    dp[j]=(p[i]==s[j] || p[i]=='?') ? dp[j+1] : false;
            }
        }
        return dp[0];
    }