Two players are playing a series of games of Rock–paper–scissors. There are a total of K games played. Player 1 has a sequence of moves denoted by string A and similarly player 2 has string B. If any player reaches the end of their string, they move back to the start of the string. The task is to count the number of games won by each of the player when exactly K games are being played.
Input: k = 4, a = “SR”, b = “R”
Output: 0 2
Game 1: Player1 = S, Player2 = R, Winner = Player2
Game 2: Player1 = R, Player2 = R, Winner = Draw
Game 3: Player1 = S, Player2 = R, Winner = Player2
Game 4: Player1 = R, Player2 = R, Winner = Draw
Input: k = 3, a = “S”, b = “SSS”
Output: 0 0
All the games are draws.
Approach: Let length of string a be n and length of string b be m. The observation here is that the games would repeat after n * m moves. So, we can simulate the process for n * m games and then count the number of times it gets repeated. For the remaining games, we can again simulate the process since it would be now smaller than n * m. For example, in the first example above, n = 2 and m = 1. So, the games will repeat after every n * m = 2 * 1 = 2 moves i.e. (Player2, Draw), (Player2, Draw), ….., (Player2, Draw).
Below is the implementation of the above approach:
Time Complexity: O(N * M)
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Improved By : chitranayal