Smallest lexicographical rotation of a string using suffix arrays in O(n)

A suffix array is a space-efficient datastructure, which, if stored together with the original string, provides the same functions as a suffix tree.

So yes, you can think of a suffix array as a storage mechanism for suffix trees. Depending on the details, there may be performance costs to using arrays over full-blown trees, which typically are outweighed by the space-use benefits. I believe the arrays are constructed almost identically to the way trees are constructed, and, in practice, suffix arrays are the preferred method for storing/representing a suffix tree.

I think you can improve this by looking at the problem as a directed graph of pairs of positions.

For this example I will use the line with values -9, -6, -1, 3, and 5.

Because it's too hard to draw a graph with just text, I'm going to represent the pairs as a table. We can think of the cells as representing the state where all containers between those two positions have been collected. Each cell has two values, one representing the cost to go left, the other representing the cost to go right (to the next container).

The values can simply be calculated as the distance between the two points multiplied by the number of barrels outside of those two points + 1. Cells where both numbers have the same sign represent cases when all barrels of the opposite sign have been collected. These are calculated using only the number of barrels in the direction away from zero.

In the table values of X mean that you can't go in that direction (all the barrels in that direction have been taken). The rows represent the current position of the collector and the columns represent the location of the next opposite barrel.

    +------+------+------+------+------+
    |  -9  |  -6  |  -1  |   3  |   5  |
+---+------+------+------+------+------+
|-9 |      |      |      | X, 24| X, 14|
+---+------+------+------+------+------+
|-6 | 3, X |      |      | 9, 27| 6, 22|
+---+------+------+------+------+------+
|-1 |      |10, X |      |20, 8 |15, 18|
+---+------+------+------+------+------+
| 3 |24, 4 |27, 6 |16, 8 |      | X, 2 |
+---+------+------+------+------+------+
| 5 |14, X |22, X |18, X |      |      |
+---+------+------+------+------+------+

The rules for moving between squares can be somewhat confusing.

For negative numbered columns, the rules are as follows:

• Going right moves down one cell.
• Going left moves down one cell and then mirrors across the diagonal.
• If the right value is X, going left moves up to the diagonal (where the column and row are equal) and left by one.

For positive numbered columns, the rules are as follows:

• Going left moves up one cell.
• Going right moves up one cell and then mirrors across the diagonal.
• If the left value is X, going right moves down to the diagonal and right by one.

Now we can run Dijkstra's algorithm to calculate the best path, using these movement rules to traverse the graph. Our starting positions are (-1, 3) and (3, -1) with initial costs of 5 and 15, respectively. Once we've calculated values for the two end positions (left of (-9, -9) and right of (5, 5)) the smaller of the two is our answer.

The running time for each of the steps are:

• O(N log N) for initially sorting the input values along the line
• O(N^2) for calculating the table/graph
• O(N^2 log N) for running Dijkstra's on the graph (Note: there are at most two edges for any given vertex).

The first two steps are dominated by the last, so your overall runtime is O(N^2 log N) which should be a good enough runtime for the challenge.