Leetcode #2852: Sum of Remoteness of All Cells
In this guide, we solve Leetcode #2852 Sum of Remoteness of All Cells in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Union Find, Array, Hash Table, Matrix
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def sumRemoteness(self, grid: List[List[int]]) -> int:
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
n = len(grid)
dirs = (-1, 0, 1, 0, -1)
cnt = sum(x > 0 for row in grid for x in row)
ans = 0
for i, row in enumerate(grid):
for j, x in enumerate(row):
if x > 0:
s, t = dfs(i, j)
ans += (cnt - t) * s
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.