2577. Minimum Time to Visit a Cell In a Grid
Description
You are given a m x n
matrix grid
consisting of non-negative integers where grid[row][col]
represents the minimum time required to be able to visit the cell (row, col)
, which means you can visit the cell (row, col)
only when the time you visit it is greater than or equal to grid[row][col]
.
You are standing in the top-left cell of the matrix in the 0th
second, and you must move to any adjacent cell in the four directions: up, down, left, and right. Each move you make takes 1 second.
Return the minimum time required in which you can visit the bottom-right cell of the matrix. If you cannot visit the bottom-right cell, then return -1
.
Example 1:
Input: grid = [[0,1,3,2],[5,1,2,5],[4,3,8,6]] Output: 7 Explanation: One of the paths that we can take is the following: - at t = 0, we are on the cell (0,0). - at t = 1, we move to the cell (0,1). It is possible because grid[0][1] <= 1. - at t = 2, we move to the cell (1,1). It is possible because grid[1][1] <= 2. - at t = 3, we move to the cell (1,2). It is possible because grid[1][2] <= 3. - at t = 4, we move to the cell (1,1). It is possible because grid[1][1] <= 4. - at t = 5, we move to the cell (1,2). It is possible because grid[1][2] <= 5. - at t = 6, we move to the cell (1,3). It is possible because grid[1][3] <= 6. - at t = 7, we move to the cell (2,3). It is possible because grid[2][3] <= 7. The final time is 7. It can be shown that it is the minimum time possible.
Example 2:
Input: grid = [[0,2,4],[3,2,1],[1,0,4]] Output: -1 Explanation: There is no path from the top left to the bottom-right cell.
Constraints:
m == grid.length
n == grid[i].length
2 <= m, n <= 1000
4 <= m * n <= 105
0 <= grid[i][j] <= 105
grid[0][0] == 0
Solutions
Solution 1: Shortest Path + Priority Queue (Min Heap)
We observe that if we cannot move at the cell $(0, 0)$, i.e., $grid[0][1] > 1$ and $grid[1][0] > 1$, then we cannot move at the cell $(0, 0)$ anymore, and we should return $-1$. For other cases, we can move.
Next, we define $dist[i][j]$ to represent the earliest arrival time at $(i, j)$. Initially, $dist[0][0] = 0$, and the $dist$ of other positions are all initialized to $\infty$.
We use a priority queue (min heap) to maintain the cells that can currently move. The elements in the priority queue are $(dist[i][j], i, j)$, i.e., $(dist[i][j], i, j)$ represents the earliest arrival time at $(i, j)$.
Each time we take out the cell $(t, i, j)$ that can arrive the earliest from the priority queue. If $(i, j)$ is $(m - 1, n - 1)$, then we directly return $t$. Otherwise, we traverse the four adjacent cells $(x, y)$ of $(i, j)$, which are up, down, left, and right. If $t + 1 < grid[x][y]$, then the time $nt = grid[x][y] + (grid[x][y] - (t + 1)) \bmod 2$ to move to $(x, y)$. At this time, we can repeatedly move to extend the time to no less than $grid[x][y]$, depending on the parity of the distance between $t + 1$ and $grid[x][y]$. Otherwise, the time $nt = t + 1$ to move to $(x, y)$. If $nt < dist[x][y]$, then we update $dist[x][y] = nt$, and add $(nt, x, y)$ to the priority queue.
The time complexity is $O(m \times n \times \log (m \times n))$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the grid, respectively.
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