How to java Program to Check if an array contains a given two different values?

Question

I want to write a code to check the existence of given two values in a List.

List<Tag> tags = new ArrayList<>();

The requirement is to return true only if the List tag contains both "start" and "end" values. My code is like this, but it doesn't cater to the requirement.

public static boolean checkStartAndEndTimeTag(List<Tag> tags) {
        boolean isSuccess = false;
        int count = 0;
        for (Tag tag : tags) {
            if (tag.getKey().equals("start") || tag.getKey().equals("end")) {
                count++;
                if (count == 2)
                    break;
                isSuccess = true;
            }

        }
        return isSuccess;

Can someone help me with to resolve this issue?

MadProgrammer · Accepted Answer · 2020-01-17 06:12:18Z

This...

if (count == 2)
  break;
isSuccess = true;

doesn't make sense. This will set isSuccess even if there is only one match

The long winded approach

Okay, let's assuming for a second that you only care if there is at least one start and one end (discounting duplicates). One approach would be to use to state flags, one for start and one for end. To keep it simple, they would start of as 0 but would only ever be a maximum of 1 (because we don't want duplicates), then you might be able to do something like...

public static boolean checkStartAndEndTimeTag(List<Tag> tags) {
    boolean isSuccess = false;
    int starts = 0;
    int ends = 0;
    for (Tag tag : tags) {
        if (tag.getKey().equals("start")) {
            starts = 1;
        } else if (tag.getKey().equals("end")) {
            ends = 1;
        }
    }
    isSuccess = (starts + ends) == 2;
    return isSuccess;
}

Ok, you don't need isSuccess = (starts + ends) == 2; and could simply return the result of the comparison. You could also break out of the loop if (starts + ends) == 2 and save yourself from unnecessary computation

for (Tag tag : tags) {
    if (tag.getKey().equals("start")) {
        starts = 1;
    } else if (tag.getKey().equals("end")) {
        ends = 1;
    }

    if ((starts + ends) == 2) {
        break;
    }
}

Using streams...

One approach might be to make use the streams support and simply filter the List and count the results, for example...

import java.util.ArrayList;
import java.util.List;

public class Test {

    public static void main(String[] args) {
        new Test();
    }

    public Test() {
        List<Tag> tags = new ArrayList<Tag>(25);
        tags.add(new Tag("begin"));
        tags.add(new Tag("front"));
        tags.add(new Tag("start"));
        tags.add(new Tag("finish"));
        tags.add(new Tag("tail"));
        tags.add(new Tag("end"));

        boolean isSuccessful = tags.stream().filter(tag -> tag.getKey().equals("start") || tag.getKey().equals("end")).count() >= 2;
        System.out.println(isSuccessful);
    }

    public class Tag {
        private String key;

        public Tag(String key) {
            this.key = key;
        }

        public String getKey() {
            return key;
        }


    }
}

Updated...

Okay, this got complicated fast. Let's assume you don't want to match two start tags, so you MUST have both one end and one start tag

So, using the above, example, we can modify the Tag class to support equals (and by extension hashcode)

public class Tag {

    private String key;

    public Tag(String key) {
        this.key = key;
    }

    public String getKey() {
        return key;
    }

    @Override
    public String toString() {
        return getKey();
    }

    @Override
    public int hashCode() {
        int hash = 7;
        hash = 73 * hash + Objects.hashCode(this.key);
        return hash;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj) {
            return true;
        }
        if (obj == null) {
            return false;
        }
        if (getClass() != obj.getClass()) {
            return false;
        }
        final Tag other = (Tag) obj;
        if (!Objects.equals(this.key, other.key)) {
            return false;
        }
        return true;
    }

}

Then we can simply use distinct to filter out any duplicates, for example...

boolean isSuccessful = tags
        .stream()
        .distinct()
        .filter(tag -> tag.getKey().equals("start") || tag.getKey().equals("end"))
        .count() >= 2;

Probably not the most efficient solution, but certainly one of the shortest

Scary Wombat · Accepted Answer · 2020-01-17 05:37:17Z

2

In this code

if (tag.getKey().equals("start") || tag.getKey().equals("end")) {
     count++;
     if (count == 2)
            break;
     isSuccess = true;
}

you are setting isSuccess to true whenever the tag is start or end.

Better way would be

if (tag.getKey().equals("start") || tag.getKey().equals("end")) {
    count++;
    if (count == 2)
        return true;
}

answered Jan 17, 2020 at 5:37

Scary Wombat

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4 Comments

daniu Over a year ago

That would erroneously return true if the array contains "start" twice.

daniu Over a year ago

I just realized that mistake was in the original code to begin with...

Scary Wombat Over a year ago

@daniu Furthermore it erroneously return true if the array contains "end" twice - yeah I noticed this when writing the question but forgot to mention.

Indi Over a year ago

Okay! It works fine. I set isSuccess = true inside if loop and return isSuccess out of the for loop. Thankyou!

Hai Hoang · Accepted Answer · 2020-01-17 06:20:27Z

2

You could also use

tags.stream()
    .map(Tag::getKey)
    .distinct()
    .filter(t -> "start".equals(t) || "end".equals(t))
    .count() == 2;

This would also fix the issue that your original code falsely returns true if the list contains statt or end twice.

edited Jan 17, 2020 at 6:20

Hai Hoang

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answered Jan 17, 2020 at 5:54

daniu

15.1k7 gold badges37 silver badges64 bronze badges

6 Comments

MadProgrammer Over a year ago

Interesting. I tried your code against the test data from my answer ([begin, front, start, finish, tail, start]) and it returns true ... I might be doing something wrong though and it returns false for [begin, front, start, finish, tail, start, end]

MadProgrammer Over a year ago

Oddly, System.out.println("Map = " + tags.stream().map(Tag::getKey).collect(Collectors.toList())); will print Map = [begin, front, start, finish, tail, start, end] which doesn't seem to be removing the duplicates - maybe it's my equals/hashcode methods?

Hai Hoang Over a year ago

You should call distinct after map tho remove the duplicate strings

Hai Hoang Over a year ago

if the list after map contains 3 or more duplicate of either "start" or "end" and not the other, using > 2 still return true.

daniu Over a year ago

It's Friday and it seems to show for us all 😁 accepted @HaiHoang s edit with distinct.

|

G.B. · Accepted Answer · 2020-01-17 06:16:00Z

-2

What about

tags.containsAll(Arrays.asList(new Tag("start"), new Tag("stop")))

edited Jan 17, 2020 at 6:16

answered Jan 17, 2020 at 5:44

G.B.

112 bronze badges

1 Comment

MadProgrammer Over a year ago

(Not a downvoted) The original List doesn't contain Strings, it contains a Tag object - whose implementation we don't know

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