Problem Description

In this problem, we have a grid in the form of an n x n matrix, where each cell can hold a value of 0, 1, or -1, each representing an empty cell, a cherry, or a thorn respectively. Our goal is to collect as many cherries as possible by following these rules:

1. Start at the top-left corner of the grid (0, 0) and move to the bottom-right corner (n - 1, n - 1) only moving right or down.
2. After reaching (n - 1, n - 1), return to (0, 0) by moving left or up.
3. While passing through a cell that contains a cherry (1), collect it, turning the cell into an empty cell (0).
4. If there's a thorn (-1) in a cell, you cannot pass through it.
5. If there's no valid path from (0, 0) to (n - 1, n - 1), you cannot collect any cherries.

Based on these rules, we have to determine the maximum number of cherries that can be collected.

Intuition

The solution to this problem utilizes dynamic programming. We think of two paths taken simultaneously: one going from (0, 0) to (n - 1, n - 1) and the other moving back to the start. Since both paths are traversed simultaneously and each move for both paths is either down or right, the total number of moves for both paths will be the same at every step. We use k to represent the total number of steps taken so far by each path.

We can represent the state of our traversal using a 3D DP array dp[k][i1][i2], where k is the total number of steps each path has taken, i1 is the row of the first path, and i2 is the row of the second path. By knowing the row of each path and the total number of steps, we can easily determine the column for each path (j1 and j2) since j1 = k - i1 and j2 = k - i2.

At every step, we attempt to transition from the previous state by deciding from which previous cell each path could have come from: either from the left or from above for each path. This gives us four possible previous states to consider. We calculate the maximum number of cherries collected among these four possibilities while ensuring that we don't step on cells with thorns or move outside the grid and we update the current state with the maximum cherries collected.

Note that we only add the cherries from grid cell grid[i1][j1] once if both paths are at the same cell since one cherry cannot be picked twice.

In the end, the maximum number of cherries collected will be the maximum value contained in the DP array for the state representing both paths returning to (0, 0).

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a 3D dynamic programming (DP) approach. The key idea is to create a DP table that will keep track of the maximum cherries collected at every step for every position of two simultaneous paths. The DP table is represented as dp[k][i1][i2], where k is the total number of steps taken, i1 is the row position in the grid of the first path, and i2 is the row position of the second path.

Implementation Steps:

1. Initialize a 3D list dp with dimensions 'two times n minus one' by n by n, filled with -inf to represent the worst case (not being able to reach certain positions). The reason for the first dimension to be 'two times n minus one' is because the maximum number of steps k it would take to reach from (0, 0) to (n - 1, n - 1) and back is 2n - 2, accounting for each possible step state.

2. Set the starting position dp[0][0][0] to the value of the starting cell grid[0][0], because that is where both paths start and the initial number of cherries collected.

3. Iterate over k from 1 to 2n - 2 to cover all possible step states.

4. For each k, iterate over all possible row positions i1 and i2 for both paths. We calculate the corresponding column positions j1 and j2 by k - i1 and k - i2 as per the constraint that i + j equals the steps taken since both paths started at (0, 0).

5. For each i1 and i2 pair, check if the corresponding j1, j2 are within bounds and the cells are not blocked by thorns (grid[i1][j1] != -1 and grid[i2][j2] != -1).

6. Compute the cherry count t that could be picked up by moving to this position for both paths. Add the values of grid[i1][j1] and grid[i2][j2] if the destinations are different, else add any one of them once.

7. Update the current state dp[k][i1][i2] by considering all the possible previous states from which the paths could have come: (i1 - 1, j1) and (i1, j1 - 1) for the first path, (i2 - 1, j2) and (i2, j2 - 1) for the second one. Take the maximum of the current state and the cherry count t plus the previous state value dp[k - 1][x1][x2].

8. After filling the DP table, the answer will be the value at dp[-1][-1][-1], representing the maximum cherries that can be collected by both paths after they've returned to (0, 0). If no path exists, the result should not be negative, so we return the maximum of 0 and this value.

By using dynamic programming, this approach avoids recalculating the result for overlapping subproblems, as we would in a naive recursive approach, thereby significantly reducing the time complexity.

The algorithm capitalizes on the insight that since the paths can be traversed simultaneously, we essentially are looking for the best paths from (0, 0) to (n - 1, n - 1) and back that cover the most cells containing cherries without crossing into any cells with thorns.

Example Walkthrough

Let's consider a small 3x3 grid to illustrate the solution approach:

[
 [ 1,  1, -1],
 [ 1, -1,  1],
 [ 1,  1,  1]
]

Each value represents a cell in the grid; 1 is a cherry, -1 is a thorn, and cells with value 0 would represent empty cells, though there are none in this example.

Step-by-Step Solution:

1. Initialize the DP table: We create a DP table with dimensions 5x3x3 since k can range from 0 to 2 * 3 - 2 = 4. Initially, the DP table is filled with -inf, but we set dp[0][0][0] to 1 as this is the value of the starting cell grid[0][0].

2. Iterate over all steps k: We begin with k=1 and proceed to k=4.

3. Iterate over all pairs of row positions (i1, i2):

   • For k=1, there are two possible positions for each path, (0, 1) or (1, 0) for both i1 and i2. However, due to thorns, (0, 1) isn't viable for either path. So we only consider (1, 0) (down for both paths).
   • Assume the upper path moves down to [1][0] and the lower path also moves down to [1][0]. The total cherries collected would be 1 from this step, making dp[1][1][1] equal to 2 (including the cherry from dp[0][0][0]).

4. Compute cherry count t:

   • Continuing with the steps, update dp[k][i1][i2] accordingly based on the viable previous positions (ignoring positions with thorns).
   • For example, when k=3, if one path is at [2][1] (moved right from [2][0]) and the other is at [1][2] (moved down to [2][0] and then moved right), we would have j1 = 2 and j2 = 1. This position is particularly interesting since both paths can potentially add cherries. If both paths hadn't collected the cherry at [2][0] initially, they could both collect it now.

5. Update the DP table:

   • At each step k, update the dp[k][i1][i2] entry with the maximum number of cherries collected based on the previous positions of the paths.
   • For instance, dp[3][2][1] would be based on the maximum of:
     • dp[2][1][1] (both paths moved down from [1][1] and [1][0]),
     • dp[2][1][0] + grid[2][1] (upper path moved down, lower path moved right),
     • dp[2][2][0] (invalid since j1 would be out of bounds),
     • dp[2][2][1] (both paths moved right).

6. Return the result: After we have filled up the DP table, we look at dp[-1][-1][-1], which is dp[4][2][2]. This value represents the maximum number of cherries collected by the two simultaneous paths. Since cells with thorns are non-viable, if all paths to (n - 1, n - 1) are blocked, the DP cell will be -inf, and thus we return the maximum of this value and 0.

By following this grid and tracing the steps stated above, we can visualize how the DP table is filled at each step k and how we account for paths that can't move due to thorns or have already picked cherries. The final DP table will give us the maximum number of cherries (in viable scenarios) that we could collect by traversing the grid twice using two paths according to the given rules.

Solution Implementation

Time and Space Complexity

The given Python code represents a dynamic programming approach to solve the problem of picking cherries in a grid, where two persons start from the top-left corner and move to the bottom-right corner collecting cherries and avoiding thorns.

Time Complexity:

The time complexity is determined by the number of states in the dynamic programming matrix and the number of operations needed to compute each state. There are 3 nested loops:

1. The outer k loop which goes from 1 to 2n - 2. This gives us a O(2n - 1) complexity factor.
2. The two inner i1 and i2 loops, each ranging from 0 to n - 1. This gives us a O(n^2) complexity factor.
3. Finally, the two most inner loops iterate over x1 and x2 which have a constant range of -1 to 1. This results in a constant time complexity, which in this case can be considered O(1) since it doesn't scale with n.

Multiplying these together gives us a time complexity of O((2n - 1) * n^2 * 1), which simplifies to O(2n^3 - n^2). Since we're interested in big-O notation, we drop the constant coefficients and lower order terms, simplifying further to O(n^3).

Space Complexity:

The space complexity is determined by the size of the dp array. The dp array has dimensions len(grid) << 1 - 1 by n by n, which results in space complexity of O((2n - 1) * n^2). This simplifies to O(2n^3 - n^2). With big-O notation, we simplify this further to O(n^3) since the squared term is negligible compared to the cubic term for large n.

In conclusion, both the time and space complexities are O(n^3).

Learn more about how to find time and space complexity quickly using problem constraints.