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994. Rotting Oranges 👍

Approach 1: Brute Force

  • Time: $O(kmn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int orangesRotting(vector<vector<int>>& grid) {
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = grid.size();
    const int n = grid[0].size();

    auto isNeighborRotten = [&](int i, int j, const vector<vector<int>>& grid) {
      for (const auto& [dx, dy] : dirs) {
        const int r = i + dx;
        const int c = j + dy;
        if (r < 0 || r == m || c < 0 || c == n)
          continue;
        if (grid[r][c] == 2)
          return true;
      }
      return false;
    };

    int ans = 0;

    while (true) {
      vector<vector<int>> nextGrid(m, vector<int>(n));
      // Calculate `nextGrid` based on `grid`.
      for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j)
          if (grid[i][j] == 1) {  // fresh
            // Any of the 4-directionally oranges is rotten
            if (isNeighborRotten(i, j, grid))
              nextGrid[i][j] = 2;
            else
              nextGrid[i][j] = 1;
          } else if (grid[i][j] == 2) {  // rotten
            nextGrid[i][j] = 2;          // Keep rotten.
          }
      if (nextGrid == grid)
        break;
      grid = nextGrid;
      ++ans;
    }

    return any_of(grid.begin(), grid.end(),
                  [&](vector<int>& row) {
      return ranges::any_of(row, [&](int orange) { return orange == 1; });
    })
               ? -1
               : ans;
  }
};

Approach 2: BFS

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int orangesRotting(vector<vector<int>>& grid) {
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = grid.size();
    const int n = grid[0].size();
    int countFresh = 0;
    queue<pair<int, int>> q;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1)
          ++countFresh;
        else if (grid[i][j] == 2)
          q.emplace(i, j);

    if (countFresh == 0)
      return 0;

    int step = 0;
    for (; !q.empty(); ++step)
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [i, j] = q.front();
        q.pop();
        for (const auto& [dx, dy] : dirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          if (grid[x][y] != 1)
            continue;
          grid[x][y] = 2;   // Mark grid[x][y] as rotten.
          q.emplace(x, y);  // Push the newly rotten orange to the queue.
          --countFresh;     // Decrease the count of fresh oranges by 1.
        }
      }

    return countFresh == 0 ? step - 1 : -1;
  }
};