Removing from a Set and Looping Over It in Java

HashSet<String> colors = new HashSet<>(List.of("red", "blue", "green"));

colors.remove("blue");

colors.remove("pink");          // not in it, nothing happens

for (String color : colors) {
    System.out.println(color);  // red and green, in any order
}

Full lesson:

s.remove(x) removes x from the set s. It returns true if x was there. If it wasn't, nothing happens and it returns false.

On a set, remove always takes the value. A set has no indexes. Careful with a List<Integer>, though. There, remove(2) removes the item at index 2.

for (String x : s) visits each value of the set once. The order isn't fixed. It may differ from the order you added the values in.

So loop over a set when the order doesn't matter. A total, a count and a check are fine.

Don't remove values from a set while you loop over it. Java stops with a ConcurrentModificationException. Remove first, then loop.

Your task

In a word game, the set found holds the words a player found. The words "qqq" and "zzz" aren't allowed. Remove both from found.

The player never found "zzz", so removing it changes nothing.

Then add up the points. A word scores one point per letter. Loop over found and add each word's length to score.

The code already prints how many words are left, and the points.

Expected output:

4 words
14 points