Comparing if a string's character exist in an array of characters

Question

I want the user to input a string, then I want to check if each charachter in this string exists in an array of charachters I created. Even if it's not in the correct order.

The way I go about it is initialise the array of chars then through using the scanner have a String input from the user.

public static char[]aa={'A','C','D','E','F','G','H','I','K','L','M','N','P','Q','R','S','T','V','W','Y','U','O','B','J','Z','X'};

I created a function

private static void isValidSequence(String sequence, char[] k) {
    outter :for (int j = 0; j < sequence.length(); j++) {
        for (int i = 0; i < k.length; i++) {
            if(sequence.charAt(j) == k[i]){
                break;
            } else {     
                System.out.println("invalid"); 
                break outter;
            }
       }     
    }
}

What happens is that if for example the first letter of the the string doesn't match the first input of array it gives me an 'invalid' input. How can I go around that? and make it iterate through the whole array of characters before giving the invalid output.

Possible duplicate of Compare 2 strings in Java upto a delimiter — Scrobot
– Scrobot, Commented Apr 30, 2019 at 8:23

Lino · Accepted Answer · 2019-04-30 08:55:30Z

1

An approach would be to sort your array, and then use the Binary Search Algorithm (BSA):

// sort the array once
Arrays.sort(aa);

// iterate over the input string
for(int i = 0, length = sequence.length(); i < length; i++) {
    // the java implementation of the BSA returns negative numbers for not found elements
    if(Arrays.binarySearch(aa, sequence.charAt(i)) < 0) {
        // char was not found, break loop
        return;
    }
}

Note: If the array is not sorted / can not be sorted, then the BSA is useless and will produce undefined results.

Note 2: The BSA is faster (O(log n)) than simple iteration (O(n))

edited Apr 30, 2019 at 8:55

answered Apr 30, 2019 at 8:50

Lino

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Comments

nullPointer · Accepted Answer · 2019-04-30 08:34:52Z

0

This can also be done as below :

char[]aa={'A','C','D','E','F','G','H','I','K','L','M','N','P','Q','R','S','T','V','W','Y','U','O','B','J','Z','X'};

String s = "aStrinG";

for (char c : s.toCharArray()) {
    for (char c2 : aa) {
       if (c2 == c) {
          System.out.println("String char " + c + " exists in char array");
       }
    }
}

It produces :

String char S exists in char array
String char G exists in char array

edited Apr 30, 2019 at 8:34

answered Apr 30, 2019 at 8:26

nullPointer

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Comments

Gurwinder Singh · Accepted Answer · 2019-04-30 08:41:11Z

0

Best way is to use SET instead of an Array. A set contains no duplicate elements and then simply you can use it using the method contains()

Using Java 9 (Unmodifiable Sets)

Set<Character> alphabetSet = Set.of('A', 'B', 'C');

//Another way
//Set<Character> alphabetSet = new HashSet<>(Arrays.asList('A', 'B', 'C'));

for(char c: sequence)
{
    if(alphabetSet.contains(c)){
      //do something
    }
    else{
      //do something
    }
}

Read more about Set: https://docs.oracle.com/javase/7/docs/api/java/util/Set.html

answered Apr 30, 2019 at 8:41

Gurwinder Singh

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Comments

anthony yaghi · Accepted Answer · 2019-04-30 08:50:47Z

0

To directly answer your question: Solution 1: Use continue instead of break this will print "invalid" each time a character is not in k.

Solution 2: You can use a counter and increment it every time you have a character not in k, then outside the loop you will have visibility on how many invalid characters you have.

Solution 3: If you want even more detail you can have a list of characters and add each invalid character to this list. This will give you visibility on exactly what characters are invalid.

I am not sure what you are trying to do so I don't know which method is better for you, you can also use an altogether approach using streams for example.

answered Apr 30, 2019 at 8:50

anthony yaghi

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2 Comments

LearningHowToCodePleaseHelpMe Over a year ago

I don't want it to print invalid everytime, I want to print invalid on the first character that doesn't match and exit the function. is that possible?

anthony yaghi Over a year ago

replace first break with continue, replace break outter with return

AxelH · Accepted Answer · 2019-04-30 09:53:22Z

If you start using the Collection provided, you could use a Set to do your checks.

First, convert the array into the Set :

char[] array = "abcdefghijklmnopqrstuvwxyz".toCharArray();

Set<Character> setAllowed = new TreeSet<>();
for(char c : array){
    setAllowed.add(c);
}

Then, you just have to iterate and check for each character. I would add another Set to retrieve every characters not allowed, giving a better output.

Set<Character> setError = new TreeSet<>();
for(char c : s.toCharArray()){
    if(setAllowed.contains(c)){
        setError.add(c);
    }
}

Test :

public static void main(String[] args) {
    String s = "foobar123";
    char[] array = "abcdefghijklmnopqrstuvwxyz".toCharArray();

    //INIT
    Set<Character> setAllowed = new TreeSet<>();
    Set<Character> setError = new TreeSet<>();
    for(char c : array){
        setAllowed .add(c);
    }

    //RESEARCH
    for(char c : s.toCharArray()){
        if(setAllowed.contains(c)){
            setError.add(c);
        }
    }

    //OUTPUT
    System.out.println(setError);
}

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