Problem Description

You have an n × n matrix called grid where each cell contains either:

• A positive integer value, or
• -1 which represents a blocked cell

You can move between non-blocked cells that share an edge (up, down, left, right).

The remoteness R[i][j] for each cell (i, j) is defined as:

• If cell (i, j) is non-blocked: R[i][j] equals the sum of all values in non-blocked cells that cannot reach cell (i, j) through any path
• If cell (i, j) is blocked (value is -1): R[i][j] = 0

Your task is to calculate and return the sum of R[i][j] for all cells in the grid.

Example understanding:

• If the grid has multiple disconnected groups of non-blocked cells (separated by blocked cells), then for any cell in one group, its remoteness would be the sum of all values in the other unreachable groups
• Cells within the same connected component can reach each other, so they don't contribute to each other's remoteness

The solution uses DFS to identify connected components. For each component with sum s and size t, every cell in that component contributes s to the remoteness of all (cnt - t) cells in other components, where cnt is the total number of non-blocked cells in the grid.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The grid represents a graph where non-blocked cells are nodes and edges exist between adjacent non-blocked cells that share an edge.

Is it a tree?

• No: The graph is not necessarily a tree. It can have multiple disconnected components (groups of connected non-blocked cells separated by blocked cells), and within each component, there could be cycles.

Is the problem related to directed acyclic graphs?

• No: The problem involves an undirected graph (you can move between adjacent cells in any direction).

Is the problem related to shortest paths?

• No: The problem is about finding connected components and calculating sums based on unreachable cells, not about finding shortest paths.

Does the problem involve connectivity?

• Yes: The core of the problem is identifying which cells are connected (can reach each other) and which cells are in separate components (cannot reach each other).

Disjoint Set Union

• While DSU could work here, the solution uses DFS because we need to:

  1. Find all connected components
  2. Calculate the sum of values in each component
  3. Count the number of cells in each component

  DFS naturally allows us to traverse each component while collecting these statistics.

Conclusion: The flowchart leads us to use DFS for this connectivity problem. DFS helps us identify connected components in the grid, calculate the sum of values in each component, and determine which cells cannot reach each other (are in different components) to compute the remoteness values.

Intuition

The key insight is recognizing that cells can only contribute to each other's remoteness if they cannot reach each other. This happens when cells belong to different connected components separated by blocked cells (-1).

Think of the grid as islands of connected non-blocked cells separated by water (blocked cells). Each island is isolated from the others. The remoteness of a cell is essentially asking: "What is the sum of all cell values on other islands that I cannot reach?"

Let's break down the thought process:

1. Connected Components: First, we need to identify which cells belong together. Cells that can reach each other form a connected component. We can use DFS to explore each component starting from any unvisited non-blocked cell.

2. Component Statistics: For each connected component, we need two pieces of information:

   • The sum s of all cell values in this component
   • The number of cells t in this component

3. Calculating Remoteness: Here's the clever part - instead of calculating remoteness for each cell individually, we can think about it from a component perspective:

   • If a component has sum s and contains t cells
   • There are cnt - t cells in other components (where cnt is the total number of non-blocked cells)
   • Each of those cnt - t cells will have s added to their remoteness (since they cannot reach any cell in this component)
   • So this component contributes s × (cnt - t) to the total remoteness

4. Final Answer: By processing each connected component once and calculating its contribution to the total remoteness, we avoid redundant calculations. The sum of all these contributions gives us the final answer.

This approach is efficient because instead of checking reachability between every pair of cells, we leverage the fact that reachability is transitive within components and impossible between different components.

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

Solution Approach

The implementation uses DFS to identify and process each connected component in the grid. Here's how the solution works:

1. Count Total Non-blocked Cells

cnt = sum(x > 0 for row in grid for x in row)

First, we count all non-blocked cells (positive values) in the entire grid. This gives us the total number of cells that can potentially contribute to remoteness calculations.

2. DFS Function for Component Exploration

def dfs(i: int, j: int) -> (int, int):
    s, t = grid[i][j], 1
    grid[i][j] = 0
    for a, b in pairwise(dirs):
        x, y = i + a, j + b
        if 0 <= x < n and 0 <= y < n and grid[x][y] > 0:
            s1, t1 = dfs(x, y)
            s, t = s + s1, t + t1
    return s, t

The DFS function explores a connected component starting from cell (i, j):

• s accumulates the sum of all cell values in the component (initialized with current cell's value)
• t counts the number of cells in the component (initialized to 1)
• We mark visited cells by setting them to 0 to avoid revisiting
• We explore all 4 directions using dirs = (-1, 0, 1, 0, -1) with pairwise to get direction pairs
• For each valid unvisited neighbor, we recursively call DFS and accumulate the results

3. Process Each Component

for i, row in enumerate(grid):
    for j, x in enumerate(row):
        if x > 0:
            s, t = dfs(i, j)
            ans += (cnt - t) * s

We iterate through the grid to find unvisited non-blocked cells:

• When we find one, it's the start of a new connected component
• We use DFS to get the component's sum s and size t
• The contribution to total remoteness is (cnt - t) * s:
  • (cnt - t) is the number of cells in other components
  • Each of these cells has s added to their remoteness
  • This avoids double counting since we process each component only once

4. Key Optimization Instead of calculating remoteness for each cell individually, we calculate each component's contribution to the total remoteness in one go. This reduces the time complexity from O(n⁴) (checking every pair of cells) to O(n²) (visiting each cell once).

The algorithm cleverly uses the grid itself to track visited cells by setting them to 0, avoiding the need for a separate visited array.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider this 3×3 grid:

[ 2, -1,  3]
[ 4,  5, -1]
[-1,  6,  7]

Step 1: Count total non-blocked cells

• Non-blocked cells (positive values): 2, 3, 4, 5, 6, 7
• Total count cnt = 6

Step 2: Identify connected components using DFS

Starting from top-left, we scan the grid:

• Cell (0,0) has value 2 > 0, start DFS:

  • Visit (0,0): value = 2
  • Check neighbors: only (1,0) is reachable
  • Visit (1,0): value = 4
  • Check neighbors: only (1,1) is reachable
  • Visit (1,1): value = 5
  • No more reachable neighbors
  • Component 1: sum = 2+4+5 = 11, size = 3 cells

• Continue scanning, cell (0,2) has value 3 > 0, start DFS:

  • Visit (0,2): value = 3
  • No reachable neighbors (blocked by -1)
  • Component 2: sum = 3, size = 1 cell

• Continue scanning, cell (2,1) has value 6 > 0, start DFS:

  • Visit (2,1): value = 6
  • Check neighbors: only (2,2) is reachable
  • Visit (2,2): value = 7
  • No more reachable neighbors
  • Component 3: sum = 6+7 = 13, size = 2 cells

Step 3: Calculate total remoteness

For each component, calculate its contribution to total remoteness:

• Component 1 (sum=11, size=3):

  • Cells in other components: cnt - size = 6 - 3 = 3
  • Contribution: 11 × 3 = 33
  • (Each of the 3 cells in other components has 11 added to their remoteness)

• Component 2 (sum=3, size=1):

  • Cells in other components: 6 - 1 = 5
  • Contribution: 3 × 5 = 15
  • (Each of the 5 cells in other components has 3 added to their remoteness)

• Component 3 (sum=13, size=2):

  • Cells in other components: 6 - 2 = 4
  • Contribution: 13 × 4 = 52
  • (Each of the 4 cells in other components has 13 added to their remoteness)

Total remoteness = 33 + 15 + 52 = 100

Verification (optional): Let's verify by calculating individual cell remoteness:

• Cells in Component 1 (2,4,5): Each has remoteness = 3 + 13 = 16
• Cell in Component 2 (3): Has remoteness = 11 + 13 = 24
• Cells in Component 3 (6,7): Each has remoteness = 11 + 3 = 14
• Total = 3×16 + 1×24 + 2×14 = 48 + 24 + 28 = 100 ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The algorithm iterates through each cell in the n × n grid exactly once in the main loop. For each cell that contains a positive value, it initiates a DFS traversal. The DFS function visits each cell in a connected component exactly once (marking visited cells by setting them to 0). Since each cell is visited at most once throughout the entire execution (either skipped if already 0 or visited once during DFS), the total number of operations is proportional to the number of cells in the grid, which is n².

Space Complexity: O(n²)

The space complexity consists of two main components:

1. Recursion stack space: In the worst case, the DFS could traverse all cells in the grid if they form a single connected component. This would create a recursion depth of O(n²) in the call stack.
2. Input grid modification: The algorithm modifies the input grid in-place by setting visited cells to 0, but this doesn't add extra space beyond the input.

Therefore, the dominant factor is the recursion stack, giving us a space complexity of O(n²).

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

Common Pitfalls

1. Modifying the Original Grid

The most significant pitfall in this solution is that it modifies the input grid by setting visited cells to 0. This destructive approach can cause issues if:

• The grid needs to be preserved for other operations
• The function is called multiple times on the same grid
• The problem explicitly requires maintaining the original input

Solution: Use a separate visited tracking mechanism instead of modifying the grid:

def sumRemoteness(self, grid: List[List[int]]) -> int:
    grid_size = len(grid)
    visited = [[False] * grid_size for _ in range(grid_size)]
  
    def dfs(row: int, col: int) -> tuple[int, int]:
        visited[row][col] = True
        component_sum = grid[row][col]
        component_count = 1
      
        for dr, dc in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
            next_row, next_col = row + dr, col + dc
            if (0 <= next_row < grid_size and 
                0 <= next_col < grid_size and 
                not visited[next_row][next_col] and
                grid[next_row][next_col] > 0):
                adjacent_sum, adjacent_count = dfs(next_row, next_col)
                component_sum += adjacent_sum
                component_count += adjacent_count
      
        return component_sum, component_count

2. Confusing Direction Array Usage

The direction array [-1, 0, 1, 0, -1] with index-based access can be confusing and error-prone. Developers might incorrectly access indices or misunderstand the pattern.

Solution: Use explicit direction pairs for clarity:

directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]  # right, down, left, up

for dr, dc in directions:
    next_row = row + dr
    next_col = col + dc

3. Integer Overflow for Large Grids

For very large grids with high values, the multiplication (cnt - t) * s could potentially overflow in languages with fixed integer sizes. While Python handles arbitrary precision integers, this could be an issue when porting to other languages.

Solution: Be aware of the maximum possible values and use appropriate data types:

• For a grid of size n×n with maximum value V, the worst case is approximately n² × n² × V
• In other languages, consider using long/int64 types
• Add validation for extremely large inputs if necessary

4. Incorrect Handling of Edge Cases

The code doesn't explicitly handle edge cases like:

• Empty grid (n = 0)
• Grid with all blocked cells (-1 values)
• Grid with no blocked cells

Solution: Add explicit edge case handling:

def sumRemoteness(self, grid: List[List[int]]) -> int:
    if not grid or not grid[0]:
        return 0
  
    grid_size = len(grid)
    total_positive_cells = sum(cell > 0 for row in grid for cell in row)
  
    # If no positive cells or all cells are connected, remoteness is 0
    if total_positive_cells == 0:
        return 0

5. Misunderstanding the Problem Definition

Developers might incorrectly calculate remoteness by:

• Including cells from the same component
• Double-counting contributions
• Misinterpreting what "cannot reach" means

Solution: Add clear documentation and consider breaking down the calculation:

# For each component, calculate its contribution to total remoteness:
# - Component with sum S and size T
# - There are (total_cells - T) cells in other components
# - Each of these cells has S added to their remoteness
# - Total contribution: (total_cells - T) * S