We are given a maze which is a cube of NxNxN cells. We start at (1,1,1) and we move to (N,N,N), in each move visiting one of the three adjacents cells in the positive x, y or z directions. Each cell of the maze contains either a '(', ')' or '.'. A path is the sequence of cells visited during the journey. We intend to find the number of such paths that produce a valid parenthesized expression.
Periods can occur freely in an expression, and must be ignored while checking if an expression is properly paranthesized or not. A paranthesized expression is said to be valid only if it adheres to the following grammar (quotes for clarity): <expr> ::= empty-string | "("<expr>")" | <expr>"." | <expr><expr>
e.g., "(())", "....", ".()." and "().(.)" are valid parathesized expressions. The maze is given in a String[] which, when concantenated, represents the cube encoded in the following manner. The characters of maze correspond to the following cube coordinates, in order: (1,1,1), (1,1,2), ..., (1,1,N), (1,2,1), (1,2,2), ..., (1,N,N), (2,1,1), ..., (N,N,N)
Given the String[] maze return an int representing the number of possible paths forming valid parenthesized expressions. If there are more than 1,000,000,000 paths, return -1 instead.
Definition
Class:
BracketMaze
Method:
properPaths
Parameters:
String[], int
Returns:
int
Method signature:
int properPaths(String[] maze, int N)
(be sure your method is public)
Constraints
-
maze will contain between 1 and 50 elements, inclusive.
-
Each element of maze will contain between 1 and 50 characters, inclusive.
-
The maze when concatenated will contain exactly N^3 characters.
-
Each character in maze will be '(', ')', or '.'.
-
N will be between 1 and 13, inclusive.
Examples
0)
{"()()()()"}
2
Returns: 2
There are two paths in this 2x2x2 cube. From the 3 possible moves in the beginning, you cannot go right directly which completes a "()" but all it's neighbours are ')', hence lead to an invalid expression. The other two possible moves lead to unique valid paths.
1)
{")()()()("}
2
Returns: 0
2)
{ "..........................." }
3
Returns: 90
3)
{"..(....)"}
2
Returns: 2
4)
{ "(.........................)"}
3
Returns: 90
import java.util.Arrays;
public class BracketMaze {
String s = "";
int n;
long[][][][] memo;
long dfs(int x, int y, int z, int p) {
char c = s.charAt(x * n * n + y * n + z);
if (x==y && y==z && z==n-1)
return (p==0 && c=='.' || p==1 && c==')') ? 1 : 0;
if (memo[x][y][z][p] != -1)
return memo[x][y][z][p];
long ret = 0;
int p1 = p;
if (c == '(')
p1++;
else if (c == ')')
p1--;
if (p1 >= 0) {
if (x < n - 1)
ret += dfs(x + 1, y, z, p1);
if (y < n - 1)
ret += dfs(x, y + 1, z, p1);
if (z < n - 1)
ret += dfs(x, y, z + 1, p1);
}
return memo[x][y][z][p] = ret;
}
public int properPaths(String[] maze, int N) {
for (int i = 0; i < maze.length; i++)
s += maze[i];
n = N;
memo = new long[N][N][N][3*N];
for (long[][][] mmm : memo)
for (long[][] mm : mmm)
for (long[] m : mm)
Arrays.fill(m, -1);
long count = dfs(0, 0, 0, 0);
return (count > 1e9) ? -1 : (int) count;
}
}