Is Sum of Remoteness of All Cells easy or hard?

The problem is typically classified as Medium. The challenge is recognizing that remoteness depends on connected components rather than individual cells, allowing the solution to be optimized from repeated traversals to a single O(m*n) pass.

Sum of Remoteness of All Cells Python/Java solution

Most implementations use DFS on the grid. Traverse neighbors in four directions, accumulate the component sum and size, then apply the formula componentSize * (totalSum - componentSum). The same logic translates directly to Python, Java, C++, Go, or TypeScript.

How to solve Sum of Remoteness of All Cells in O(n)?

Treat the grid as a graph and compute connected components. First calculate the sum of all non-blocked cells. Run DFS to determine the size and value sum of each component. For a component with sum S and size K, add K * (totalSum - S) to the answer. This visits every cell once, giving O(m*n) time.

What is the best approach for Sum of Remoteness of All Cells?

The optimal approach groups cells into connected components using DFS. Compute the total sum of all valid cells in the grid, then for each component calculate its sum and size. Each cell contributes totalSum minus componentSum to the remoteness, so the component contributes size * (totalSum - componentSum). This processes the grid in O(m*n) time.

Is Sum of Remoteness of All Cells asked at Google/Amazon/Meta?

Grid traversal and connected component problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants involving DFS, BFS, or Union Find on matrices are common because they test graph modeling and optimization skills.

What data structure is used in Sum of Remoteness of All Cells?

The primary data structures are a grid traversal stack or recursion for DFS, a visited matrix to avoid revisiting cells, and optionally a Union Find structure to track connected components. These help group reachable cells efficiently.

What is the time complexity of Sum of Remoteness of All Cells?

Using the DFS connected component method, the time complexity is O(m*n) because each cell in the grid is visited once. The space complexity is also O(m*n) due to the visited structure and recursion stack or explicit stack.

Description Hints 3 Solution Video 12:46

2852. Sum of Remoteness of All Cells

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You are given a 0-indexed matrix grid of order n * n. Each cell in this matrix has a value grid[i][j], which is either a positive integer or -1 representing a blocked cell.

You can move from a non-blocked cell to any non-blocked cell that shares an edge.

For any cell (i, j), we represent its remoteness as R[i][j] which is defined as the following:

If the cell (i, j) is a non-blocked cell, R[i][j] is the sum of the values grid[x][y] such that there is no path from the non-blocked cell (x, y) to the cell (i, j).
For blocked cells, R[i][j] == 0.

Return the sum of R[i][j] over all cells.

Example 1:

Input: grid = [[-1,1,-1],[5,-1,4],[-1,3,-1]]
Output: 39
Explanation: In the picture above, there are four grids. The top-left grid contains the initial values in the grid. Blocked cells are colored black, and other cells get their values as it is in the input. In the top-right grid, you can see the value of R[i][j] for all cells. So the answer would be the sum of them. That is: 0 + 12 + 0 + 8 + 0 + 9 + 0 + 10 + 0 = 39.
Let's jump on the bottom-left grid in the above picture and calculate R[0][1] (the target cell is colored green). We should sum up the value of cells that can't be reached by the cell (0, 1). These cells are colored yellow in this grid. So R[0][1] = 5 + 4 + 3 = 12.
Now let's jump on the bottom-right grid in the above picture and calculate R[1][2] (the target cell is colored green). We should sum up the value of cells that can't be reached by the cell (1, 2). These cells are colored yellow in this grid. So R[1][2] = 1 + 5 + 3 = 9.

Example 2:

Input: grid = [[-1,3,4],[-1,-1,-1],[3,-1,-1]]
Output: 13
Explanation: In the picture above, there are four grids. The top-left grid contains the initial values in the grid. Blocked cells are colored black, and other cells get their values as it is in the input. In the top-right grid, you can see the value of R[i][j] for all cells. So the answer would be the sum of them. That is: 3 + 3 + 0 + 0 + 0 + 0 + 7 + 0 + 0 = 13.
Let's jump on the bottom-left grid in the above picture and calculate R[0][2] (the target cell is colored green). We should sum up the value of cells that can't be reached by the cell (0, 2). This cell is colored yellow in this grid. So R[0][2] = 3.
Now let's jump on the bottom-right grid in the above picture and calculate R[2][0] (the target cell is colored green). We should sum up the value of cells that can't be reached by the cell (2, 0). These cells are colored yellow in this grid. So R[2][0] = 3 + 4 = 7.

Example 3:

Input: grid = [[1]]
Output: 0
Explanation: Since there are no other cells than (0, 0), R[0][0] is equal to 0. So the sum of R[i][j] over all cells would be 0.

Constraints:

1 <= n <= 300
1 <= grid[i][j] <= 10⁶ or grid[i][j] == -1

Asked at 1 company

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Topics

Hints open one at a time

Hint 1

We consider each cell of grid a node on a graph, and there is an edge between two nodes if one of them doesn't contain -1 and they share an edge on the grid.

Hint 2

We find the connected components of said graph.

Hint 3

The nodes of each connected component have the same value for R and it's the sum of the values of all nodes in other connected components.

12:46

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Examples

Case 1Case 2Case 3

Input

[[-1,1,-1],[5,-1,4],[-1,3,-1]]

Input

[[-1,3,4],[-1,-1,-1],[3,-1,-1]]

Input

[[1]]

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