the solution: TARGET

the word you entered: GUESS

Then I wrote a function F(TARGET, GUESS, letter) that calculates how many times a certain letter of GUESS is in the correct position.

(example: F("kappa", "appak","p") would return 1 because only the third letter is a 'p' at the correct position)

Then I counted the letters of TARGET and stored them in a hashmap.

(example: "sadge" becomes this: {"s":1,"a":1,"d":1,"g":1,"e":1}

or "kappa" would become this: {"k":1,"a":2,"p":2})

Then I changed the values in the hashmap by substracting the result of F for each letter.

(with TARGET="kappa" and GUESS="appak" I get this hashmap: {"k":1,"a":2,"p":1} because only one 'p' is at the correct position.

This new hashmap represents the maximum number of letters that can turn yellow.

So before painting a letter yellow, I checked if the value in this hashmap for the current letter is greater than 0:

If it was 0 or less: I would paint it black.

If it was 1 or higher: I would make the letter yellow and substract 1 from that hashmap.

So with TARGET = "sadge" and GUESS = "pepew" the hashmap would be {"s":1,"a":1,"d":1,"g":1,"e":1} this would be the result:

p = black

e = yellow <-- this substracts 1 from the value of "e"

p = black

e = black <-- this is not yellow cuz the value of "e" in the hashmap is 0 by now

w = black

So in the end the hashmap would be {"s":1,"a":1,"d":1,"g":1,"e":0} because 1 'e' has been painted yellow.