lesson17_1_NumberSolitaire.rb

# A game for one player is played on a board consisting of N consecutive squares, numbered from 0 to N − 1. There is a number written on each square. A non-empty array A of N integers contains the numbers written on the squares. Moreover, some squares can be marked during the game.

# At the beginning of the game, there is a pebble on square number 0 and this is the only square on the board which is marked. The goal of the game is to move the pebble to square number N − 1.

# During each turn we throw a six-sided die, with numbers from 1 to 6 on its faces, and consider the number K, which shows on the upper face after the die comes to rest. Then we move the pebble standing on square number I to square number I + K, providing that square number I + K exists. If square number I + K does not exist, we throw the die again until we obtain a valid move. Finally, we mark square number I + K.

# After the game finishes (when the pebble is standing on square number N − 1), we calculate the result. The result of the game is the sum of the numbers written on all marked squares.

# For example, given the following array:

#     A[0] = 1
#     A[1] = -2
#     A[2] = 0
#     A[3] = 9
#     A[4] = -1
#     A[5] = -2
# one possible game could be as follows:

# the pebble is on square number 0, which is marked;
# we throw 3; the pebble moves from square number 0 to square number 3; we mark square number 3;
# we throw 5; the pebble does not move, since there is no square number 8 on the board;
# we throw 2; the pebble moves to square number 5; we mark this square and the game ends.
# The marked squares are 0, 3 and 5, so the result of the game is 1 + 9 + (−2) = 8. This is the maximal possible result that can be achieved on this board.

# Write a function:

# def solution(a)

# that, given a non-empty array A of N integers, returns the maximal result that can be achieved on the board represented by array A.

# For example, given the array

#     A[0] = 1
#     A[1] = -2
#     A[2] = 0
#     A[3] = 9
#     A[4] = -1
#     A[5] = -2
# the function should return 8, as explained above.

# Write an efficient algorithm for the following assumptions:

# N is an integer within the range [2..100,000];
# each element of array A is an integer within the range [−10,000..10,000].

def solution(a)
  ([a.first]+Array.new(a.size-1){ -Float::INFINITY}).tap do |dp|
    (a.size - 1).times do |i| 
      1.upto(6) { |die| a[i+die] ? dp[i+die] = [dp[i+die],dp[i]+a[i+die] ].max : break }
    end
  end.last
end