Problem Description

In this problem, we are given an n x n matrix called grid, where the value at each cell in the grid represents either a positive integer or -1 indicating a blocked cell. The aim is to calculate the sum of "remoteness" for each cell in the grid. For a non-blocked cell (i, j), the remoteness R[i][j] is defined as the sum of values of all non-blocked cells (x, y) from which there is no path to cell (i, j). For blocked cells (where the value is -1), the remoteness is 0.

To visualize how to move within the grid, one can move horizontally or vertically but not diagonally to any adjacent non-blocked cell. The overall goal is to determine the sum of remoteness for all cells in the matrix and return this sum.

Flowchart Walkthrough

First, let's navigate the algorithm using the Flowchart. Here's a detailed analysis:

Is it a graph?

• Yes: The grid is a graph where each cell corresponds to a node, and edges exist between adjacent cells.

Is it a tree?

• No: A grid typically has multiple possible paths between nodes and lacks the hierarchical structure of a tree.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem deals with a grid, which does not define any directed or acyclic constraints.

Is the problem related to shortest paths?

• Yes: The problem involves calculating remoteness based on distances, which is inherently a shortest path problem.

Is the graph weighted?

• No: The problem likely describes a uniform grid where each edge has the same weight.

Does the problem involve connectivity?

• Yes: We need to calculate the remoteness for all cells, implying a connectivity issue across the graph.

Conclusion: Based on the analysis using the Flowchart and the characteristics of the problem, Depth-First Search (DFS) can be effectively used to explore the connectivity and paths in the unweighted grid to compute remoteness due to its ability to thoroughly explore all potential paths and connections.

Intuition

To approach this problem, we have to understand that the remoteness of a non-blocked cell depends on the inaccessibility of other cells to this particular cell. Hence, if we know all the sets of cells that cannot reach a particular cell, we can calculate that cell's remoteness by summing up the values of those inaccessible cells.

The intuition behind the solution uses Depth-First Search (DFS) to group cells into connected components. For a non-blocked cell, we perform DFS to find all other reachable non-blocked cells. During each DFS, we keep track of:

1. The sum s of the values of the cells visited in the current DFS.
2. The total count t of cells visited in the current DFS.

Each time we visit a cell, we mark it as part of a connected component by setting its value to 0 to avoid revisiting it. Once the DFS is complete for a connected component, we know that:

• Any cell not visited during this DFS cannot reach any cell in the connected component, and their values should contribute to the remoteness of the component.
• The sum s represents the sum of values within the connected component.
• The count t represents the total number of cells in the connected component.

The remoteness for each non-blocked cell within the same connected component is the same. Hence, the remoteness (R[i][j]) for any non-blocked cell (i, j) is the product of the difference between the total count of non-blocked cells and the size of the connected component t (i.e., (cnt - t)) and the sum s of the connected component.

The final answer is the sum over the remoteness of all cells. By iterating over each cell and performing the above steps for each cell if it's a non-blocked cell that hasn't been visited, we can find the sum of remoteness for all cells in the grid.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The solution implements Depth-First Search (DFS) to walk through the grid and find connected components of non-blocked cells. Here's a step-by-step explanation of the implementation:

1. Define the dfs function that takes the current indices i, j as parameters. This function returns a tuple (s, t), where s is the sum of the values of the cells within the current connected component and t is the number of cells in this connected component.

2. The dfs function first initializes s with the current cell's value and t with 1 (since the current cell is being counted). It then marks the current cell as part of a connected component by setting grid[i][j] to 0, preventing the algorithm from revisiting this cell in the future.

3. The function proceeds to iterate over all adjacent cells—in the up, down, left, and right directions—checking if any of them are non-blocked (where grid[x][y] > 0). For each non-blocked adjacent cell, it recursively calls dfs to visit it.

4. The results of the recursive calls (s1, t1) are aggregated to the current DFS call, updating the s and t for the entire connected component.

5. The solution records the total count of non-blocked cells in the cnt variable before starting the DFS processes. This count is needed to compute the remoteness for the cells.

6. The solution iterates over all the cells using nested loops. If a cell has a value greater than 0 (non-blocked and not yet visited), it calls the dfs function on it.

7. For each DFS completed, it has the sum s and total t for the connected component. The algorithm computes the remoteness for the cells in this component as (cnt - t) * s and adds it to the answer ans. This calculation is based on the count of cells not in the connected component (cnt - t) and the sum of the values of cells within the component.

8. Finally, after traversing all cells, the ans variable holds the sum of the remoteness of all cells which the function returns.

The implementation makes use of basic Python constructs like list comprehensions to calculate the count of non-blocked cells and for-loop along with tuple unpacking for iteration. The dfs function facilitates the use of recursion as a primary strategy to search through the grid, exhibiting the DFS pattern.

In summary, the algorithm complexity is driven by the DFS pattern, which ensures that each cell is explored only once. The data structures used are simple: the matrix grid itself, to mark visited cells, and a couple of variables to keep track of the summation. By leveraging these data structures and algorithm patterns, the implementation efficiently solves the problem.

Example Walkthrough

Initial setup:

Let's consider a small 3 x 3 grid as our example:

grid = [
    [1, 2, -1],
    [-1, 3, 4],
    [5, -1, 6]
]

In this case, we have blocked cells at positions (0,2), (1,0), and (2,1).

Step-by-step DFS analysis:

1. Start at cell (0,0) with a value of 1. Initialize s to 1 and t to 1. We perform a DFS from this cell, marking it as visited by setting its value to 0.

• No valid adjacent cells to the right (blocked) or down (blocked).
• DFS ends for this cell with s=1, t=1.

2. Move to cell (0,1) with a value of 2. This cell starts a new DFS as it is not visited yet.

• It can move down to (1,1) with a value of 3, where s = 2+3=5, t=1+1=2.
• From (1,1), it moves right to (1,2) with a value of 4, where s = 5+4=9, t = 2+1=3.
• All adjacent cells are visited/blocked, so DFS ends with s=9, t=3.

3. Skipping to cell (2,0) with a value of 5. This cell is not visited yet.

• No valid adjacent cells to move to.
• DFS ends for this cell with s=5, t=1.

4. Checking the last cell (2,2) with a value of 6.

• No valid adjacent cells to move to.
• DFS ends for this cell with s=6, t=1.

Calculating remoteness:

The total count of non-blocked cells cnt is 6.

For cell (0,0), with s=1, t=1:

• Remoteness R for this connected component is (cnt - t) * s = (6 - 1) * 1 = 5.

For cells (0,1), (1,1), and (1,2), with s=9, t=3:

• Remoteness R for this connected component is (cnt - t) * s = (6 - 3) * 9 = 27.

For cell (2,0), with s=5, t=1:

• Remoteness R for this connected component is (cnt - t) * s = (6 - 1) * 5 = 25.

For cell (2,2), with s=6, t=1:

• Remoteness R for this connected component is (cnt - t) * s = (6 - 1) * 6 = 30.

Final remoteness sum:

Adding the remoteness of all cells: 5 + 27 + 25 + 30 = 87.

So, the sum of remoteness for this 3 x 3 grid example is 87.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python code performs a depth-first search (DFS) to compute the "sum of remoteness" on a 2D grid. To analyze the time complexity, we consider the following points:

• There is a nested for-loop that iterates through each element of the n x n grid, which gives a time complexity of O(n^2) for this part.
• Within the loop, the code executes a DFS (the dfs function) whenever a positive grid value is encountered. In the worst case, DFS may visit each cell once.
• The time complexity of DFS in a grid is proportional to the number of cells in the grid, which is O(n^2) for an n x n grid.
• Since DFS is called once per positive number in the grid and ensures that each cell is visited only once (by marking visited cells with 0), the entire grid is processed only once. Thus, the combined complexity of processing the grid with DFS is still O(n^2).

Given these considerations, the overall time complexity of the function is O(n^2).

Space Complexity

For the space complexity:

• The space is dominated by the recursion stack of DFS since we don't use any additional data structures that are dependent on n.
• In the worst case, when the grid is filled completely with positive numbers, the DFS could go as deep as O(n^2) since in the worst case it might have to visit all cells before backtracking (for example, in a scenario where you have a snake-like pattern).
• Apart from the recursion stack, the dfs function contains variable declarations which use only constant extra space.

Hence, the space complexity of the function is O(n^2).

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