I am trying to solve a Codility problem for the dynamic programming section provided.
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:
class Solution { public int solution(int[] 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].
I get low performance with the online judge.
My solution is provided below.
public static int solution(int[] A) {
int N = A.length;
int[] result = new int[N];
result[0] = A[0];
for (int i = 1; i < N; i++) {
result[i] = result[i - 1];
for (int j = 2; j <= 6; j++) {
if (j > i) {
break;
}
// 0, 1, 2, 3, 4
result[i] = Math.max(result[i], result[j - 2]);
}
result[i] += A[i];
}
return result[N - 1];
}
How do I improve the correctness and performance?
You can do it in O(N)
with O(N)
additional space.
Allocate array dp
of length N
and for every next item in the array find out the biggest possible value of the game by following formula:
dp[k] = max(dp[k-6], dp[k-5], .., dp[k - 1]) + data[k]
Some special handling is needed in the beginning of the array. The answer will be stored in dp[N-1]
.
EDIT: As properly noted @גלעד ברקן in the comment, we can do it with O(1)
space. To do this we can use circular buffer of length 7.
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