Java maps, sets, and duplicate counting

Track counts and distinct values with separate collection roles.

Java

import java.util.Map;
import java.util.TreeMap;
import java.util.Set;
import java.util.TreeSet;
class Main {
    public static void main(String[] args) {
        Map<String, Integer> counts = new TreeMap<>();
        Set<String> unique = new TreeSet<>();
        for (String word : new String[]{"red", "blue", "red"}) {
            counts.put(word, counts.getOrDefault(word, 0) + 1);
            unique.add(word);
        }
        System.out.println(counts);
        System.out.println(unique.size());
    }
}

Output

{blue=1, red=2}
2

Explanation

A Map associates keys with values; a Set records distinct values. getOrDefault avoids a missing-key null result while counting.

TreeMap provides sorted key order, and TreeSet provides sorted unique values. Using these ordered collections makes the displayed result deterministic.

Explore the dry run

Guided walkthroughReady

1import java.util.Map;

2import java.util.TreeMap;

3import java.util.Set;

4import java.util.TreeSet;

5class Main {

6 public static void main(String[] args) {

7 Map<String, Integer> counts = new TreeMap<>();

8 Set<String> unique = new TreeSet<>();

9 for (String word : new String[]{"red", "blue", "red"}) {

10 counts.put(word, counts.getOrDefault(word, 0) + 1);

11 unique.add(word);

12 }

13 System.out.println(counts);

14 System.out.println(unique.size());

15 }

16}

Follow the example

Start to inspect the source, values, and output. These prepared steps explain the published sample; grouped loop steps describe multiple iterations.

View the complete dry-run table

Prepared trace of the published program. Some rows group repeated iterations.
What happens	State after step	Output so far
Create an ordered map.	`counts = {}`	—
Process red, blue, and red, incrementing existing counts.	`counts = {blue=1, red=2}`	—
The repeated red does not create a third distinct entry.	`unique = [blue, red]`	—
Print in sorted key order.	`unique = [blue, red]`	{blue=1, red=2}
Print the distinct count.	`unique = [blue, red]`	{blue=1, red=2} 2

Cost and common mistakes

O(n log k) comparisons for n words and k distinct keys, ignoring variable string comparison cost; O(k) storage.

HashMap does not promise the same iteration order.
An object's ordering/equality rules must be suitable for its use as a key.

Try it yourself

Add another blue token.

Reveal answer and reasoning

The map becomes {blue=2, red=2}; the distinct count remains 2.

Java maps, sets, and duplicate counting

Explanation

Follow the example

Cost and common mistakes

Try it yourself

Further reading