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Problem solution · C++

Sum of Remoteness of All Cells

Sum of Remoteness of All Cells: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Sum of Remoteness of All Cells, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 49 lines of C++ from the credited upstream file 2852.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSum of Remoteness of All Cells · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long sumRemoteness(vector<vector<int>>& grid) {    const long sum = accumulate(grid.begin(), grid.end(), 0L,                                [](long acc, const vector<int>& row) {      return acc +             accumulate(row.begin(), row.end(), 0L,                        [](long acc, int num) { return acc + max(0, num); });    });     long ans = 0;     for (int i = 0; i < grid.size(); ++i)      for (int j = 0; j < grid[0].size(); ++j)        if (grid[i][j] > 0) {          const auto [count, componentSum] = dfs(grid, i, j);          ans += (sum - componentSum) * count;        }     return ans;  }  private:  static constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};   // Returns the (count, componentSum) of the connected component that contains  // (x, y).  pair<int, long> dfs(vector<vector<int>>& grid, int i, int j) {    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())      return {0, 0};    if (grid[i][j] == -1)      return {0, 0};     int count = 1;    long componentSum = grid[i][j];    grid[i][j] = -1;  // Mark as visited.;     for (const auto& [dx, dy] : kDirs) {      const int x = i + dx;      const int y = j + dy;      const auto [nextCount, nextComponentSum] = dfs(grid, x, y);      count += nextCount;      componentSum += nextComponentSum;    }     return {count, componentSum};  }};

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