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2577. Minimum Time to Visit a Cell In a Grid 👍

  • Time: $O(mn\log mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minimumTime(vector<vector<int>>& grid) {
    if (grid[0][1] > 1 && grid[1][0] > 1)
      return -1;

    const int m = grid.size();
    const int n = grid[0].size();
    const vector<int> dirs{0, 1, 0, -1, 0};
    using T = tuple<int, int, int>;  // (time, i, j)
    priority_queue<T, vector<T>, greater<>> minHeap;
    vector<vector<bool>> seen(m, vector<bool>(n));

    minHeap.emplace(0, 0, 0);
    seen[0][0] = true;

    while (!minHeap.empty()) {
      const auto [time, i, j] = minHeap.top();
      minHeap.pop();
      if (i == m - 1 && j == n - 1)
        return time;
      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 (seen[x][y])
          continue;
        const int extraWait = (grid[x][y] - time) % 2 == 0 ? 1 : 0;
        const int nextTime = max(time + 1, grid[x][y] + extraWait);
        minHeap.emplace(nextTime, x, y);
        seen[x][y] = true;
      }
    }

    throw;
  }
};

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class Solution {
  public int minimumTime(int[][] grid) {
    if (grid[0][1] > 1 && grid[1][0] > 1)
      return -1;

    final int m = grid.length;
    final int n = grid[0].length;
    final int[] dirs = {0, 1, 0, -1, 0};
    // (time, i, j)
    Queue<int[]> minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
    boolean[][] seen = new boolean[m][n];

    minHeap.offer(new int[] {0, 0, 0});
    seen[0][0] = true;

    while (!minHeap.isEmpty()) {
      final int time = minHeap.peek()[0];
      final int i = minHeap.peek()[1];
      final int j = minHeap.poll()[2];
      if (i == m - 1 && j == n - 1)
        return time;
      for (int k = 0; k < 4; ++k) {
        final int x = i + dirs[k];
        final int y = j + dirs[k + 1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        if (seen[x][y])
          continue;
        final int extraWait = (grid[x][y] - time) % 2 == 0 ? 1 : 0;
        final int nextTime = Math.max(time + 1, grid[x][y] + extraWait);
        minHeap.offer(new int[] {nextTime, x, y});
        seen[x][y] = true;
      }
    }

    throw new IllegalArgumentException();
  }
}

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class Solution:
  def minimumTime(self, grid: List[List[int]]) -> int:
    if grid[0][1] > 1 and grid[1][0] > 1:
      return -1

    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]
    minHeap = [(0, 0, 0)]  # (time, i, j)
    seen = {(0, 0)}

    while minHeap:
      time, i, j = heapq.heappop(minHeap)
      if i == m - 1 and j == n - 1:
        return time
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if (x, y) in seen:
          continue
        extraWait = 1 if (grid[x][y] - time) % 2 == 0 else 0
        nextTime = max(time + 1, grid[x][y] + extraWait)
        heapq.heappush(minHeap, (nextTime, x, y))
        seen.add((x, y))