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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) {
    const int m = grid.size();
    const int n = grid[0].size();
    const vector<int> dirs{0, 1, 0, -1, 0};

    auto isNeighborRotten = [&](int i, int j, const vector<vector<int>>& grid) {
      for (int k = 0; k < 4; ++k) {
        const int r = i + dirs[k];
        const int c = j + dirs[k + 1];
        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
            if (isNeighborRotten(
                    i, j, grid))  // Any of 4-directionally oranges is rotten
              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 any_of(row.begin(), row.end(),
                    [&](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) {
    const int m = grid.size();
    const int n = grid[0].size();
    const vector<int> dirs{0, 1, 0, -1, 0};
    int ans = 0;
    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;

    while (!q.empty()) {
      ++ans;
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [i, j] = q.front();
        q.pop();
        for (int k = 0; k < 4; ++k) {
          const int x = i + dirs[k];
          const int y = j + dirs[k + 1];
          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 newly rotten orange to queue
          --countFresh;     // Decrease the count of fresh oranges by 1
        }
      }
    }

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