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LESSON 238 OF 677 · SETS

Removing from a Set and Looping Over It in Kotlin

Lesson

TL ; DR:

remove(x) takes x out of a set. A for loop visits every value once, in no fixed order.


val colors = hashSetOf("red", "blue", "green")
colors.remove("blue")
colors.remove("pink")           // not in it, nothing happens

for (color in colors) {
    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.

for (x in s) visits each value of the set once. The order of a HashSet 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. Kotlin 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
STEP 1 OF 3
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