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How would you go about solving this problem?

\n

You have write a method to check wether you win, lose or tie in a game of Rock-Paper-Scissor, based on the choice of you and your opponent.
\nThe method should return 2 if you win, 1 if you tie and 0 if you lose.
\nAnd we can use only two if statements and no 'or' or 'and' operators

\n

Lets use some characters and values to represent the choices and outcomes:
\nRock = A | Paper = B | Scissor = C
\nWin = 6 | Tie = 3 | Loss = 0

\n

Here is the program you need to make changes to, replace the places where there's ... with only a single line of your code.

\n
RockPaperScissor.java
\n
public class RockPaperScissor {\n    ...\n    public static int checkWinner(char player, char opponent) {\n        if(...) return 3;\n        ...\n        if(...) return 6;\n        return 0;\n    }\n}
\n

Try to find the solution yourself before you read the next section.

\n
\n
\nNeed a hint?\n
RockPaperScissor.java
\n
public class RockPaperScissor {\n    //Use an array of characters to represent the choices    static char[] choices = {'C', 'A', 'B', 'C'};    public static int checkWinner(char player, char opponent) {\n        if(...) return 3;\n        ...\n        if(...) return 6;\n        return 0;\n    }\n}
\n
\n
\n
\nNeed one more hint?\n

Look at this pattern
\nWhenever the player wins,
\nthese are the plays:
\nplayer: A beats C :opponent
\nplayer: B beats A :opponent
\nplayer: C beats B :opponent

\n

If we organize this a bit differently, we can find a pattern
\nplayer choice
\nA B C A
\nC A B C <- the array we created
\nopponent choice

\n

Notice how whenever we win, the opponent's choice is always one value behind of the player's choice in the array.
\nie.
\nwhen the player choice is B, the opponent's choice is A
\nwhen the player choice is A, the opponent's choice is C, because C is one value behind A in the array.
\nthis is the extra element at the start.

\n
\n
\n
\nHere's my solution to the problem-\n
RockPaperScissor.java
\n
public class RockPaperScissor {\n    static char[] choices = {'C', 'A', 'B', 'C'};\n    public static int checkWinner(char player, char opponent) {\n        //In case of a tie\n        if(player == opponent) return 3;\n        //In case of a win\n        if(choices[player - 'A'] == opponent)  return 6;\n        //In case of a loss\n        return 0;\n    }\n}
\n

Okay, so what did I do here?
\nI used the choices array to store the choices.
\nI used the player - 'A' + 1 to get the index of the player's choice in the choices array.
\nHow does that work you ask?
\nWell, the ASCII value of A is 65 and the ASCII value of C is 67.
\nSo, when the player gives a character input from A to C, we actually have a value from 65 to 67.
\nIn order to find this value in the choices array, we need to subtract 65 from the player input.
\nSo we get a range from 0-2.
\nBut the choices array starts from 1 because of the extra elements on both the ends.
\nSo we add 1 to the value.

\n

Now since we know that we can only win if the opponent's choice is one value behind the player's choice, we can use the choices array to check if the player wins or not.
\nSo to check for the win, we can simply use choices[player - 'A' + 1 - 1] == opponent (simplified to choices[player - 'A'] == opponent)

\n

And that's how you solve this in just three lines of code./\nThanks for reading :))))))

\n
\n
\n

This is from one of the problems in the Advent of Code 2022.

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