Category: LeetCode Programming

Difficulty: Hard

Topics: Array Breadth-First Search Graph Heap (Priority Queue) Matrix Shortest Path

#91 🗺️ 2577. Minimum Time to Visit a Cell In a Grid 🚀🧠

Imagine you are navigating a grid 🚦. Each cell in this grid has a “minimum unlock time,” meaning you can only visit it after waiting a certain amount of time⏳. Starting from the top-left cell, your goal is to reach the bottom-right cell as quickly as possible while adhering to the timing rules. If no valid path exists, return -1.

📝 Problem Statement

You are given an m x n grid where:

grid[row][col] represents the minimum time required to visit cell (row, col).
You start at the top-left corner (0, 0) at t = 0.
You can move up, down, left, or right, and each move takes 1 second.

Your task is to return the minimum time required to reach the bottom-right corner of the grid, or -1 if it’s impossible.

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

🧪 Examples

Example 1:

Input:

grid = [[0,1,3,2],[5,1,2,5],[4,3,8,6]]

Output:

Explanation:
One of the valid paths to reach (2,3) is:

Start at (0,0) at t = 0.
Move to (0,1) at t = 1 (allowed because grid[0][1] <= 1).
Move to (1,1) at t = 2 (allowed because grid[1][1] <= 2).
Move to (1,2) at t = 3.
Move back and forth between cells to adjust timing.
Finally, move to (2,3) at t = 7.

Example 2:

Input:

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

Output:

-1

Explanation:
There is no valid path from the top-left to the bottom-right corner.

🚀 Approach to the Solution 🚀

1. Understand the Constraints

Each cell has a minimum time requirement (grid[row][col]), which dictates when the cell becomes accessible.
Movement is allowed only to the adjacent cells (up, down, left, right), and each step takes exactly 1 second.
If the current time is less than the cell’s minimum time requirement, you must wait until the cell becomes accessible.
Waiting can involve time adjustments due to parity (odd/even differences in time).

2. Represent the Problem as a Graph

The grid can be interpreted as a graph:

Each cell is a node.
Each adjacent cell is connected by an edge with a cost of 1 second plus any required waiting time.

The task is to find the shortest time to traverse this graph from the top-left corner (0, 0) to the bottom-right corner (m-1, n-1).

3. Edge Case Handling

Before starting the traversal:

Check if the top-left cell’s neighbors (i.e., (0, 1) and (1, 0)) are accessible at the start.
If both neighbors require a time greater than 1 to enter, it’s impossible to leave the starting cell, and the result is -1.

4. Algorithm Choice

Since the problem involves finding the shortest path in a weighted graph, Dijkstra’s algorithm is an ideal choice. It efficiently finds the shortest paths by exploring nodes in the order of increasing cost (time, in this case).

5. Key Steps in Dijkstra’s Algorithm

Initialization:
- Use a priority queue (min-heap) to process cells in increasing order of time.
- Maintain a distance matrix initialized to infinity (inf), where distance[row][col] represents the minimum time to reach a cell.
- Start with distance[0][0] = 0 and push (0, 0, 0) (time, row, column) into the priority queue.
Traverse the Grid:
- While the priority queue is not empty:
  - Pop the cell (time, row, col) with the smallest time.
  - For each adjacent cell (new_row, new_col):
    - Check if it is within grid bounds.
    - Calculate the time to reach the cell:
      - If the current time plus 1 (time + 1) is less than the cell’s minimum time requirement, compute the waiting time:
        
        Adjust for parity to ensure that the new time aligns with the cell’s accessibility condition.
    - If the computed time is better (smaller) than the current distance[new_row][new_col], update it and push the cell into the priority queue.
Stop When the Target is Reached:
- If the bottom-right cell (m-1, n-1) is dequeued, return its time as the result.
- If the queue is exhausted and the target is not reached, return -1.

6. Handle Waiting Time and Parity

If the current time (t) is less than grid[new_row][new_col], calculate the waiting time:

The waiting time must align the parity (odd/even) of the current time to match the cell’s accessibility time.

For example:

If grid[new_row][new_col] = 4 and t = 3, wait until t = 4.
If grid[new_row][new_col] = 5 and t = 4, wait until t = 5 (or another odd number).

🛠️ Basic Solution

We can treat the grid as a graph, where each cell is a node, and the edges represent possible moves. The challenge is that each cell has a “waiting time”⏳. If you arrive at a cell before its unlock time, you must wait until it is accessible.

We can solve this using Dijkstra’s algorithm 🐍 with a priority queue to always explore the shortest available path.

🐍 Python Implementation 🐍

Let’s dive deeper into the Python implementation of the solution. We’ll walk through the logic, the key components, and how they work together to achieve the goal of finding the minimum time to traverse the grid.

Here’s the complete implementation:

from heapq import heappush, heappop
from math import inf
from typing import List

class Solution:
    def minimumTime(self, grid: List[List[int]]) -> int:
        # Step 1: Handle edge case where starting cell neighbors are inaccessible
        if grid[0][1] > 1 and grid[1][0] > 1:
            return -1

        # Step 2: Initialize dimensions and data structures
        rows, cols = len(grid), len(grid[0])
        distance = [[inf] * cols for _ in range(rows)]
        distance[0][0] = 0  # Starting point has distance 0
        priority_queue = [(0, 0, 0)]  # (time, row, col)
        
        # Directions for movement: up, down, left, right
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        # Step 3: Implement Dijkstra's algorithm using a priority queue
        while priority_queue:
            time, row, col = heappop(priority_queue)

            # If we reach the bottom-right corner, return the minimum time
            if row == rows - 1 and col == cols - 1:
                return time

            # Explore all neighbors
            for dr, dc in directions:
                new_row, new_col = row + dr, col + dc

                # Check boundaries
                if 0 <= new_row < rows and 0 <= new_col < cols:
                    # Calculate new time to reach the neighbor
                    new_time = time + 1
                    if new_time < grid[new_row][new_col]:
                        # Adjust for waiting time based on parity
                        wait_time = (grid[new_row][new_col] - new_time) % 2
                        new_time = grid[new_row][new_col] + wait_time
                    
                    # Update distance if a shorter path is found
                    if new_time < distance[new_row][new_col]:
                        distance[new_row][new_col] = new_time
                        heappush(priority_queue, (new_time, new_row, new_col))

        # If we cannot reach the bottom-right corner, return -1
        return -1

Explanation of Code

1. Handle Edge Cases

The very first step checks whether the starting point’s neighbors are immediately inaccessible:
```
if grid[0][1] > 1 and grid[1][0] > 1:
    return -1
```
If both neighbors require more than 1 second to enter, it’s impossible to leave the starting cell, and we return -1.

2. Initialize Variables

rows and cols: Dimensions of the grid.
distance: A 2D list initialized with inf, representing the shortest time to each cell. The starting point (0, 0) is set to 0.
priority_queue: A min-heap that processes cells in order of their time. Initially, it contains only the starting point (0, 0, 0) (time, row, column).

3. Dijkstra’s Algorithm

The algorithm explores the grid while always processing the cell with the smallest time first:
```
while priority_queue:
    time, row, col = heappop(priority_queue)
```

For each cell, its neighbors are checked:

for dr, dc in directions:
    new_row, new_col = row + dr, col + dc
    if 0 <= new_row < rows and 0 <= new_col < cols:

The new time to reach the neighbor is calculated:
```
new_time = time + 1
```
If the cell’s minimum time requirement (grid[new_row][new_col]) is greater than new_time, waiting is applied:
```
wait_time = (grid[new_row][new_col] - new_time) % 2
new_time = grid[new_row][new_col] + wait_time
```

4. Update the Priority Queue

If the computed new_time is less than the previously known distance[new_row][new_col], the distance is updated, and the cell is pushed into the priority queue:

if new_time < distance[new_row][new_col]:
    distance[new_row][new_col] = new_time
    heappush(priority_queue, (new_time, new_row, new_col))

5. Return the Result

If the bottom-right cell is reached, its time is returned:

if row == rows - 1 and col == cols - 1:
    return time

If the queue is exhausted and the target cell isn’t reached, return -1:
```
return -1
```

Key Features of the Implementation

Efficient Priority Queue:
- The min-heap ensures that the cell with the smallest time is always processed first.
- This minimizes unnecessary computations.
Handling Waiting Time:
- The modulo operation aligns the waiting time to the cell’s required parity.
Boundary Checks:
- All neighbors are validated to ensure they’re within grid boundaries.
Optimal Time Complexity:
- The algorithm runs in \(O(m \cdot n \cdot \log(m \cdot n))\), making it efficient for large grids.

⚠️ Edge Cases

Starting Neighbors Blocked:
- Grid like [[0, 3], [3, 3]] → Should return -1.
Single Path:
- Grid like [[0, 1, 2, 3]] → Ensure the algorithm handles linear paths correctly.
Immediate Accessibility:
- Grid like [[0, 0], [0, 0]] → Should return the minimum time of moving directly to the target.
Waiting in Place:
- Grid like [[0, 2], [1, 3]] → Requires waiting at specific cells before advancing.

By thoroughly testing these cases, the implementation can be validated for correctness.

⏳ Time Complexity

Initialization: \(O(m \cdot n)\)
Heap Operations: \(O((m \cdot n) \cdot \log(m \cdot n))\)
Total: \(O(m \cdot n \cdot \log(m \cdot n))\)

🗄️ Space Complexity

\(O(m \cdot n)\) for the distance matrix and heap.

⚡ Optimized Solution

The basic solution already employs an optimized approach by leveraging Dijkstra’s algorithm and minimizing redundant computations.

🧮 Example Walkthrough

Let’s take a larger and more detailed example to illustrate the problem and solution step by step. This walkthrough will include specific grid configurations, the decisions made during traversal, and the logic applied at each step.

Example

Input Grid:

grid = [
    [0, 2, 4, 6],
    [3, 1, 7, 5],
    [8, 3, 2, 10],
    [9, 12, 4, 8]
]

Output: 12

Explanation: Let’s go step-by-step to find the minimum time required to traverse from the top-left cell (0,0) to the bottom-right cell (3,3).

Step 1: Starting at (0,0)

At time t=0, we are at (0,0). This cell is accessible because grid[0][0] = 0.
Push (0,0,0) into the priority queue: (time, row, col).

Step 2: Exploring Neighbors

At t=0, we process (0,0). We explore its neighbors:

Right to (0,1):
- grid[0][1] = 2 requires t ≥ 2.
- We’ll wait at (0,0) until t=2, then move to (0,1).
- Update distance for (0,1) to t=2 and push (2,0,1) into the queue.
Down to (1,0):
- grid[1][0] = 3 requires t ≥ 3.
- We’ll wait at (0,0) until t=3, then move to (1,0).
- Update distance for (1,0) to t=3 and push (3,1,0) into the queue.

Step 3: Process (0,1)

At t=2, we process (0,1). We explore its neighbors:

Right to (0,2):
- grid[0][2] = 4 requires t ≥ 4.
- We’ll wait at (0,1) until t=4, then move to (0,2).
- Update distance for (0,2) to t=4 and push (4,0,2) into the queue.
Down to (1,1):
- grid[1][1] = 1 is accessible at t=3.
- Update distance for (1,1) to t=3 and push (3,1,1) into the queue.

Step 4: Process (1,0)

At t=3, we process (1,0). We explore its neighbors:

Right to (1,1):
- Already reached with t=3. Skip.
Down to (2,0):
- grid[2][0] = 8 requires t ≥ 8.
- We’ll wait at (1,0) until t=8, then move to (2,0).
- Update distance for (2,0) to t=8 and push (8,2,0) into the queue.

Step 5: Process (1,1)

At t=3, we process (1,1). We explore its neighbors:

Right to (1,2):
- grid[1][2] = 7 requires t ≥ 7.
- We’ll wait at (1,1) until t=7, then move to (1,2).
- Update distance for (1,2) to t=7 and push (7,1,2) into the queue.
Down to (2,1):
- grid[2][1] = 3 is accessible at t=4.
- Update distance for (2,1) to t=4 and push (4,2,1) into the queue.

Step 6: Process (0,2), (2,1), and Beyond

Following the same logic:

At t=4, process (0,2):
- Move to (0,3) at t=6.
- Push (6,0,3).
At t=4, process (2,1):
- Move to (2,2) at t=5.
- Push (5,2,2).
At t=6, process (0,3):
- Move to (1,3) at t=7.
- Push (7,1,3).
At t=8, process (2,0):
- Move to (3,0) at t=9.
- Push (9,3,0).

Step 7: Reach (3,3)

Finally, at t=12, process (3,3). Since we’ve reached the bottom-right corner, return t=12.

Path Taken

The final path looks like this (with time values in parentheses):

(0,0) → (0,1)(2) → (1,1)(3) → (2,1)(4) → (2,2)(5) → (3,3)(12)

Key Observations in the Example

Waiting Periods:
- At cells where the required time is greater than the arrival time, waiting is necessary. This ensures alignment with the cell’s accessibility condition.
Optimal Movement:
- The algorithm always processes the cell with the smallest time first, ensuring that the shortest path is evaluated before longer alternatives.
Efficient Queue Management:
- Using a priority queue allows the algorithm to efficiently select the next cell to process.
Dynamic Exploration:
- All directions are explored at each step, enabling the algorithm to handle complex grids with backtracking or detours.

📜 Conclusion

This problem highlights the importance of shortest-path algorithms like Dijkstra’s for grid-based challenges, especially when additional constraints (like waiting times) are involved. Understanding how to manipulate these constraints efficiently can drastically improve performance in complex scenarios. 🧠✨

Written on November 29, 2024