Java: multiple arrays

Question

What I am using: I have multiple arrays

String[] cat = {"schema cat","schemacat","schemac"}
String[] dog = {"schema dog","schemadog","schemad"}
String[] horse = {"schema horse","schemachorse","schemah"}

I am getting a value to my method for example schema cat and I need to find in which array it is written. But I can sometimes get schemac or cat instead of schema cat.

Problem: So I need to find in which array is containing value.

What I have tried: I have tried writing multiple if cases but it's getting to messy and complicated. I am wondering what is the best solution for this case. I was thinking about switch, but don't know how to solve case situations.

• azro: Share the code of what you tried
• Yousaf: All you need to do is use a loop to iterate over each array and check if schema cat exists in any array or not.
• S.L. Barth is on codidact.com: Not entirely sure what you want... do you need to get the string if it contains a substring? Then the Java method String.contains is what you need.
• D1': I am using this to identify if I am getting cat/dog/horse. If I know which list gonna be the chosen one then I can say that it's cat or in other words get instance of cat or smth.

Answer 1

You could simple create a method, where you give the value, the different arrays to look into, and iterate over them until you found the value

Edit : to get like the name of the array, you can add it as the first value of the array and use it as return value

static String whichArrayContains(String value, String[]... arrays) {
 for (String[] array : arrays) {
 if (Arrays.asList(array).contains(value)) {
 return array[0];
 }
 }
 return null;
}

And use like

public static void main(String[] args) {
 String[] cat = {"car", "schema cat", "schemacat", "schemac"};
 String[] dog = {"dog", "schema dog", "schemadog", "schemad"};
 String[] horse = {"horse", "schema horse", "schemachorse", "schemah"};
 String value = "schema cat";
 System.out.println(Arrays.toString(whichArrayContains(value, cat, dog, horse))); // cat
 System.out.println(Arrays.toString(whichArrayContains(value, dog, horse))); //null
}

• D1': Maybe you know if I can get array name?
• azro: @D1' you cannot do that
• D1': In such case then I can add cat element to array and then I will retrieve array I can get this element and say that it's cat
• azro: @D1' for sure, you can add it as first and the method will return the first element only, i’ll edit with this

Answer 2

The simple solution would be to use lists, and a "reverse" maps, like:

Map<String, List<String>> listsByMembers = ...

The idea would be: instead of using arrays, you use Lists (just because collections work better with collections, instead of arrays). And then, for each member of each List, you do listsByMembers.put(listEntry, theList).

Then you can do a simple lookup listsByMembers.get("cat") for example.

If we are talking about a large scale solution for real world problems, that would boil down to re-invent the wheel. For real-world scenarios, you should rather look into tools like Elastic, Solr, ...

Answer 3

By using java8 streams concept you can solve your problem as shown below.

import java.util.Arrays;
public class Test {
 public static void main(String[] args) {
 String[] cat = {"schema cat","schemacat","schemac"};
 String[] dog = {"schema dog","schemadog","schemad"};
 String[] horse = {"schema horse","schemachorse","schemah"};
 String testStr = "schemac";
 System.out.println(isStringExist(testStr, cat));
 System.out.println(isStringExist(testStr, dog));
 System.out.println(isStringExist(testStr, horse));
 }
 public static boolean isStringExist(String testStr, String[] items) {
 return Arrays.stream(items).parallel().anyMatch(testStr :: contains);
 }
}

In the above code isStringExist() method accepts the String array and test string that need to be compared with each String present inside the String array items and it returns a Boolean value which confirms that the string present in the String array that is passed to the method.

• GhostCat: parallelStream()? Seriously?