Because we want to bring the size of charFrequencyMap equal to 'K' while shrinking the window. If we don't delete the leftChar from the map when its frequency is zero, our frequency map will not shrink.

what exactly this Map.put doing in this program? charFrequencyMap.put(rightChar, charFrequencyMap.getOrDefault(rightChar, 0) + 1);

Could this also be solved using a set instead of a hash map?

I have the same feeling that it can be solved with a set, but still haven't found how to to do it

dont think a set would work because you would be unable to track how many dupes

We can't use a set, as we need to keep track of the frequency of each character.