Question

I’m writing a function that returns true if the argument, which is a string, contains at least 2 characters from at least 2 categories:

• lower case letter
• upper case letter
• digit grammar
• character/everything else

EXAMPLE:

abAB => true
aB => false
ab12 => true
1ab2 => true
asdfasdf1 => false

I’m working on a regular expression that does this but am having trouble. I’ve also considered breaking up the regular expressions into multiple if statements and checking individually if at least 2 characters from each group are contained in the string.

For example:

comprised = 0
if(string contains *[0-9]*[0-9])
    comprised = comprised+1
if(string contains *[a-b]*[a-b])
    comprised = comprised+1
if(string contains *[A-Z]*[A-Z])
    comprised = comprised+1
if(string contains *[^0-9a-zA-Z]*[^0-9a-zA-Z])
    comprised = comprised+1

if comprised >= 2
    return true
else return false

Answer 1

There is really no need for regex in this situation. The java code below shows an alternative way of solving it.

public static boolean validate(String s) {
    int[] group = new int[4];
    for (char c : s.toCharArray()){
        group[getGroupForChar(c)]++;
    }
    int large = 0;
    for (int g : group){
        if (g >= 2)large++;
    }
    return large>=2;
}
private static int getGroupForChar(char c) {
    if (Character.isLowerCase(c))return 0;
    else if (Character.isUpperCase(c))return 1;
    else if (Character.isDigit(c))return 2;
    return 3;
}
public static void main(String[] args) {
    String[] tests = new String[]{
        "abAB",
        "aB",
        "ab12",
        "1ab2",
        "asdfasdf1"
    };

    for (String s : tests) {
        System.out.println(s+": "+validate(s));
    }
}

Output:

abAB: true
aB: false
ab12: true
1ab2: true
asdfasdf1: false

Site where you can test the code: fiddle