2577. Minimum Time to Visit a Cell In a Grid


Problem Description

In this problem, we're provided with a m x n 2-dimensional grid that represents a matrix of non-negative integers. Each cell in the matrix holds a value that signifies the earliest time at which that cell can be visited. The rules of movement are as follows:

  • The starting point is the top-left cell of the matrix (at time 0).
  • We can move to an adjacent cell in one of the four cardinal directions (up, down, left, or right) assuming the current time is at least the value in that cell.
  • Each move to an adjacent cell takes exactly 1 second.

The objective is to determine the minimum possible time to reach the bottom-right cell of the grid. If there's no such path that allows us to reach the bottom-right cell under the given constraints, we should return -1.

Intuition

To solve this problem, we need to carefully consider the constraints and movement rules. Since the value in each cell represents the minimum time required to enter it, we cannot simply take the shortest path; we may have to wait before entering certain cells. Given the nature of the problem, it's apparent that we need to track the passage of time during movement.

One approach to this kind of problem is to use a shortest path algorithm that's adapted to account for the time constraints in each cell. A commonly used algorithm for finding the shortest path in grid-like structures is Dijkstra's algorithm, which maintains a priority queue of nodes to be visited, ordered by the current known shortest distance (or in this case, the shortest time) to the node.

The basic idea behind the solution is to use a Min-Heap priority queue (as facilitated by Python's heapq module) to keep track of potential moves, ordered by the earliest time they can be made. The heap consists of tuples where each tuple includes the current time t at which we can enter cell (i, j), the row position i, and the column position j.

We start with the top-left cell, and at each step, we extract the cell with the earliest visitable time from the heap and consider all of its adjacent cells. If moving to an adjacent cell (x, y) from the current cell (i, j) is a valid move, we calculate the earliest visit time nt for (x, y). If nt is less than the minimum required time of cell (x, y) in the grid, we adjust nt so that it complies with the requirement of the cell. If this adjusted nt is also less than any previously known visit time for cell (x, y), it means we’ve found a faster route to this cell, and we update the time and add it to the heap to explore later.

Finally, once we reach the bottom-right cell of the matrix, we return the time t. If the heap gets empty before reaching the bottom-right cell, it means there is no valid path, and we should return -1.

Learn more about Breadth-First Search, Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The solution implements the adapted Dijkstra's shortest path algorithm using a Min-Heap to optimize for the time taken to reach each cell. Let’s walk through the implementation step-by-step:

  1. Initialization:

    • The distance matrix dist is initialized to infinity for all cells, which represents the earliest time we can get to that cell.
    • The starting cell (0, 0) is given a distance of 0, as we can begin there instantly.
    • A Min-Heap priority queue q is initialized with a tuple containing the starting cell's coordinates and the initial time 0.
  2. Exploring Cells:

    • We use a list dirs to encode the directions to move in the grid, using the pairwise pattern to represent up, down, left, and right.
    • We continuously pop cells from the Min-Heap that have the smallest current time t until it's empty or we reach the destination, (m - 1, n - 1).
  3. Visiting Adjacent Cells:

    • For each cell popped from the heap, we iterate through its valid adjacent cells using the dirs list to calculate x, y coordinates of the neighboring cells.
    • For each neighbor (x, y), we check if it's within the bounds of the grid. If it is, we calculate the next time nt we can be at cell (x, y) after moving from the current cell.
  4. Calculating Earliest Visitable Time (nt):

    • nt is the later of the current time plus 1 (since moving takes 1 second) or the minimum required time of the cell as given by the grid.
    • If the grid imposes a minimum time requirement greater than nt, we must wait to comply with it. If nt is already waiting past the grid time, we need to ensure visitation can occur at an even second, so we adjust nt accordingly with (grid[x][y] - nt) % 2.
  5. Update Distance and Heap:

    • If our calculated nt for an adjacent cell (x, y) is earlier than the previously known time in dist[x][y], it means we've found a faster route to that cell, so we update dist[x][y] to nt.
    • The updated cell with its new earliest time is then pushed onto the heap to continue the exploration process.
  6. Termination:

    • If we happen to reach the bottom-right cell (m - 1, n - 1), we immediately return the current time t, as it represents the earliest time to reach the destination.
    • If the heap gets exhausted before we reach the destination, this indicates there is no valid path, and we return -1.

The beauty of this approach is in how it consistently finds the shortest path to all cells in terms of time, considering both the grid's restrictions and the constraint that movements can only happen every second. As such, this algorithm guarantees that when we reach the destination cell, we do so in the minimum possible time.

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Example Walkthrough

Let's assume we have the following 3 x 3 grid for our problem:

1grid = [
2    [0, 1, 2],
3    [0, 1, 2],
4    [0, 1, 2]
5]

Following the approach outlined in the solution:

1. Initialization: We start by initializing a distance matrix dist - which we'll represent as part of our explanation - and a priority queue q.

  • dist starts as:
1dist = [
2    [0, ∞, ∞],
3    [∞, ∞, ∞],
4    [∞, ∞, ∞]
5]
  • The starting point (0,0) has a distance of 0.
  • q will be initialized with [(0,0,0)] which represents a tuple of the earliest time (0) and coordinates (0,0).

2. Exploring Cells: We have our dirs list as [(0,1),(1,0),(0,-1),(-1,0)] representing right, down, left, and up respectively.

3. Visiting Adjacent Cells: We will pop (0,0,0) from q and consider its neighbors (0,1) and (1,0).

4. Calculating Earliest Visitable Time (nt for (0,1)):

  • We compute nt by taking the maximum between the grid value at (0,1) which is 1 and current time + 1. Since current time is 0, we get nt as 1.
  • There's no wait time required, so this is the nt we will use.

5. Update Distance and Heap:

  • dist[0][1] is updated from to 1 and (1,0,1) is added to the heap q.

4. Calculating Earliest Visitable Time (nt for (1,0)):

  • Calculating nt for (1,0), we again compute the maximum between the grid value at (1,0) which is 0 and current time + 1 which is 1. So nt for (1, 0) is also 1.

5. Update Distance and Heap:

  • dist[1][0] is updated from to 1 and (1,1,0) is added to the heap q.

Now our dist matrix looks like this:

1dist = [
2    [0, 1, ∞],
3    [1, ∞, ∞],
4    [∞, ∞, ∞]
5]

And our heap q has [(1,0,1),(1,1,0)].

The same process continues, and we keep popping from the q moving in valid directions, updating dist if we find better nt until we reach the bottom-right cell (2, 2) or q becomes empty. Our updated dist matrix after these operations will help us to identify the minimum time to reach (2, 2).

In this small example, we eventually find that the minimum time to reach the bottom-right cell (2,2) is 4, with the dist matrix updated as:

1dist = [
2    [0, 1, 4],
3    [1, 2, 3],
4    [2, 3, 4]
5]

And hence the shortest time to reach from (0, 0) to (2, 2) according to given rules is dist[2][2], which is 4. If at any point the heap gets exhausted before we reach (2, 2), we would return -1. In our case, there is a path, so the output for this grid is 4.

Solution Implementation

1from heapq import heappop, heappush
2from itertools import pairwise
3from math import inf
4from typing import List
5
6class Solution:
7    def minimumTime(self, grid: List[List[int]]) -> int:
8        # Check for the initial impossible case where the immediate neighbors of the start are too high
9        if grid[0][1] > 1 and grid[1][0] > 1:
10            return -1
11      
12        # Get dimensions of the grid
13        rows, cols = len(grid), len(grid[0])
14      
15        # Initialize distance matrix with infinity
16        distance = [[inf] * cols for _ in range(rows)]
17        distance[0][0] = 0  # Start point has distance 0
18      
19        # Priority queue for the BFS: (time, row, col)
20        priority_queue = [(0, 0, 0)]
21      
22        # Directions for exploring neighbors (up, right, down, left)
23        directions = (-1, 0, 1, 0, -1)
24      
25        # Explore the grid until we reach the bottom-right corner
26        while priority_queue:
27            time, row, col = heappop(priority_queue)
28          
29            # If we reached the end, return the distance
30            if row == rows - 1 and col == cols - 1:
31                return time
32          
33            # Check all neighboring squares
34            for delta_row, delta_col in pairwise(directions):
35                new_row, new_col = row + delta_row, col + delta_col
36                # Ensure we are within grid boundaries
37                if 0 <= new_row < rows and 0 <= new_col < cols:
38                    new_time = time + 1  # Increment time by one step
39                    # If we reach earlier than the grid cell's value, we have to wait
40                    if new_time < grid[new_row][new_col]:
41                        new_time = grid[new_row][new_col] + (grid[new_row][new_col] - new_time) % 2
42                    # If this path is faster, update the distance and push to the queue
43                    if new_time < distance[new_row][new_col]:
44                        distance[new_row][new_col] = new_time
45                        heappush(priority_queue, (new_time, new_row, new_col))
46      
47        # If we never reached the end, we can not finish the task
48        return -1
49
1import java.util.PriorityQueue;
2import java.util.Arrays;
3
4class Solution {
5    public int minimumTime(int[][] grid) {
6        // Check for a situation where the path is blocked just next to the starting point
7        if (grid[0][1] > 1 && grid[1][0] > 1) {
8            return -1;
9        }
10
11        // m is the number of rows and n is the number of columns in the grid
12        int m = grid.length, n = grid[0].length;
13
14        // dist array holds the minimum time to reach each cell
15        int[][] dist = new int[m][n];
16        for (int[] row : dist) {
17            Arrays.fill(row, Integer.MAX_VALUE); // Fill with a large number to represent infinity
18        }
19        // Starting point has a distance of 0
20        dist[0][0] = 0;
21
22        // Priority queue for efficient retrieval of the next cell to process
23        PriorityQueue<int[]> priorityQueue = new PriorityQueue<>((a, b) -> a[0] - b[0]);
24        priorityQueue.offer(new int[]{0, 0, 0}); // Add starting cell to the queue
25
26        // Directions for moving up, left, down, right
27        int[] directions = {-1, 0, 1, 0, -1};
28
29        while (!priorityQueue.isEmpty()) {
30            int[] cell = priorityQueue.poll();
31            int time = cell[0], i = cell[1], j = cell[2];
32
33            // If we're at the bottom-right corner, return the distance
34            if (i == m - 1 && j == n - 1) {
35                return time;
36            }
37
38            // Explore all possible directions (up, left, down, right)
39            for (int k = 0; k < 4; k++) {
40                int nextX = i + directions[k];
41                int nextY = j + directions[k + 1];
42
43                // Check if the new position is within the bounds of the grid
44                if (nextX >= 0 && nextX < m && nextY >= 0 && nextY < n) {
45                    int nextTime = time + 1;
46
47                    // If waiting time is required, wait until the path is clear
48                    if (nextTime < grid[nextX][nextY]) {
49                        nextTime = grid[nextX][nextY] + (grid[nextX][nextY] - nextTime) % 2;
50                    }
51
52                    // If a shorter path to (nextX, nextY) is found, update the distance and add to the queue
53                    if (nextTime < dist[nextX][nextY]) {
54                        dist[nextX][nextY] = nextTime;
55                        priorityQueue.offer(new int[]{nextTime, nextX, nextY});
56                    }
57                }
58            }
59        }
60
61        // If we reach this point, there is no path to the bottom-right corner, return -1
62        return -1;
63    }
64}
65
1#include <vector>
2#include <queue>
3#include <tuple>
4#include <cstring>
5
6class Solution {
7public:
8    int minimumTime(vector<vector<int>>& grid) {
9        // If there is no path that can be taken right from the start, return -1.
10        if (grid[0][1] > 1 && grid[1][0] > 1) {
11            return -1;
12        }
13        int rows = grid.size(), cols = grid[0].size();
14
15        // Create a distance matrix and initialize with high values (infinite equivalent).
16        int distance[rows][cols];
17        memset(distance, 0x3f, sizeof distance);
18        // Starting point distance is 0.
19        distance[0][0] = 0;
20
21        // Use a min-heap to store the current time and positions, organized by the shortest time.
22        priority_queue<tuple<int, int, int>,
23                       vector<tuple<int, int, int>>,
24                       greater<tuple<int, int, int>>> pq;
25      
26        // Start with the position (0, 0) at time 0.
27        pq.emplace(0, 0, 0);
28
29        // Array to facilitate traversing in 4 directions: up, right, down, left.
30        int directionOffsets[5] = {-1, 0, 1, 0, -1};
31
32        // Loop until the queue is empty (it will break inside the loop on reaching the end)
33        while (!pq.empty()) {
34            auto [currentTime, row, col] = pq.top();
35            pq.pop();
36
37            // If the end is reached, return the time.
38            if (row == rows - 1 && col == cols - 1) {
39                return currentTime;
40            }
41
42            // Explore neighbors in all four directions.
43            for (int k = 0; k < 4; ++k) {
44                int newX = row + directionOffsets[k], newY = col + directionOffsets[k + 1];
45                // If the new positions are within bounds, process them.
46                if (newX >= 0 && newX < rows && newY >= 0 && newY < cols) {
47                    int newTime = currentTime + 1;
48                    // If current time is less than required time, adjust the newTime accordingly.
49                    if (newTime < grid[newX][newY]) {
50                        newTime = grid[newX][newY] + (grid[newX][newY] - newTime) % 2;
51                    }
52                    // If the calculated time is less than the known shortest time, update the distance and enqueue.
53                    if (newTime < distance[newX][newY]) {
54                        distance[newX][newY] = newTime;
55                        pq.emplace(newTime, newX, newY);
56                    }
57                }
58            }
59        }
60        // In the context of the original code, the loop is infinite; hence, we should never reach this point.
61        // We can add a return statement here for the sake of correctness, but ideally, it should never be reached.
62        return -1;
63    }
64};
65
1function minimumTime(grid: number[][]): number {
2    // Check for immediate infeasibility.
3    if (grid[0][1] > 1 && grid[1][0] > 1) {
4        return -1;
5    }
6
7    const rows = grid.length;
8    const cols = grid[0].length;
9
10    // Initialize distance matrix with high values, representing 'infinite' distance.
11    const distance: number[][] = Array.from({ length: rows }, () => 
12        Array(cols).fill(Infinity));
13
14    // Set the starting point's distance to 0.
15    distance[0][0] = 0;
16
17    // Initialize a min-heap priority queue.
18    const pq: [number, number, number][] = [[0, 0, 0]];
19
20    // Function to sort priority queue by first element (time).
21    const sortPq = () => pq.sort((a, b) => a[0] - b[0]);
22
23    // Direction offsets for facilitating traversal in 4 directions.
24    const directionOffsets = [-1, 0, 1, 0, -1];
25
26    // While there are still elements in the priority queue.
27    while (pq.length) {
28        // Get the smallest time element and remove it from the queue.
29        sortPq();
30        const [currentTime, row, col] = pq.shift()!;
31
32        // Return the time if we've reached the end.
33        if (row === rows - 1 && col === cols - 1) {
34            return currentTime;
35        }
36
37        // Explore in all four directions.
38        for (let k = 0; k < 4; k++) {
39            const newX = row + directionOffsets[k];
40            const newY = col + directionOffsets[k + 1];
41
42            // If the new positions are within bounds, process them.
43            if (newX >= 0 && newX < rows && newY >= 0 && newY < cols) {
44                let newTime = currentTime + 1;
45
46                // Adjust the newTime if earlier than required time.
47                if (newTime < grid[newX][newY]) {
48                    newTime = grid[newX][newY] + (grid[newX][newY] - newTime) % 2;
49                }
50
51                // Update the distance matrix and enqueue if found shorter path.
52                if (newTime < distance[newX][newY]) {
53                    distance[newX][newY] = newTime;
54                    pq.push([newTime, newX, newY]);
55                }
56            }
57        }
58    }
59
60    // This return statement should theoretically not be reached since the loop will exit upon reaching the end.
61    return -1;
62}
63

Time and Space Complexity

Time Complexity

The provided Python code implements a Dijkstra-like algorithm using a priority queue for finding the minimum time required to reach the bottom-right corner of the grid from the top-left corner. The main while loop runs until the destination is reached. In the worst case, this loop might run for every cell in the grid. For each cell, the loop considers at most four neighboring cells to which it may travel next.

Given m rows and n columns in the grid, there are m*n cells, and since each cell is inserted into the priority queue at most once, the complexity for all operations involving the priority queue will be O(m*n * log(m*n)).

The pairwise function used inside the loop along with the movement to (x, y) from (i, j) happens in O(1) time since the possible directions are constant.

Therefore, the overall time complexity is O(m*n * log(m*n)).

Space Complexity

The space complexity is influenced by the storage of the dist matrix, the dirs tuple, and the priority queue.

  • The dist matrix is m by n which takes O(m*n) space.
  • The dirs tuple is of constant size, so it takes O(1) space.
  • The priority queue contains at most m*n elements if all cells are pushed into it, leading to O(m*n) space.

Thus, the overall space complexity is O(m*n) for the distances plus O(m*n) for the priority queue, which simplifies to O(m*n).

Learn more about how to find time and space complexity quickly using problem constraints.


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