The Question
Coding

Minimum Time to Visit Cell in a Grid

You are given an m x n matrix grid of non-negative integers. Each grid[row][col] represents the minimum time required to enter that cell. You start at (0, 0) at time t = 0. In each second, you can move to an adjacent cell (up, down, left, right). You can only enter a cell if the current time plus one second is greater than or equal to the value at that cell. If you cannot reach a cell at the required time, you may move back and forth between previously visited cells to wait. Return the minimum time to reach the bottom-right cell (m-1, n-1). If it is impossible, return -1. Constraints: - m == grid.length, n == grid[i].length - 2 <= m, n <= 1000 - 0 <= grid[i][j] <= 10^5 - grid[0][0] == 0
Java
Dijkstra's Algorithm
PriorityQueue
Questions & Insights

Clarifying Questions

What is the range of $m$ and $n$? (Usually up to 1000 \times 1000 in competitive programming).
Can we revisit cells? Yes, the problem implies back-and-forth movement might be necessary to "wait" for a cell to become accessible.
What are the constraints on the values within `grid`? Non-negative integers, typically up to 10^5 or 10^9.
Is it possible to be stuck at the starting cell? Yes, if both (0, 1) and (1, 0) have a requirement > 1, we cannot move at t=0 and thus cannot increment time.

Assumptions

The grid size is at least 2 \times 1 or 1 \times 2.
If we can make at least one move initially, we can always "waste" time by moving back and forth between two cells.
Each move takes exactly 1 second.

Thinking Process

Shortest Path Problem: Since we want the minimum time and all "edge weights" are non-negative (time always increases), Dijkstra's algorithm is the most suitable approach.
The "Wait" Mechanic: If the target cell (nr, nc) has a requirement grid[nr][nc] > t + 1, we can't move there immediately. However, if we have an adjacent cell we can move back and forth to, we can "wait" until the time is sufficient. Each round trip takes 2 seconds, meaning we can only reach the target cell at a time T \ge grid[nr][nc] such that T has the same parity as the Manhattan distance from the start (or more simply, the same parity as t+1 plus any 2-second increments).
Parity Logic: If we need to wait until grid[nr][nc], the actual time we arrive is:
grid[nr][nc] if the difference between grid[nr][nc] and the current time t is odd.
grid[nr][nc] + 1 if the difference between grid[nr][nc] and the current time t is even.
Initial Trap: If grid[0][1] > 1 and grid[1][0] > 1, we cannot move anywhere from the start at t=0. Since we can't wait in place, we return -1.
Implementation Breakdown

Problem Set

Functional Requirements: Find the minimum time to reach (m-1, n-1) starting from (0, 0) at t=0.
Constraints:
m, n \le 1000.
grid[i][j] \le 10^5.
Move in 4 directions.

Approach

Algorithm: Dijkstra's Algorithm.
Data Structure: PriorityQueue (Min-Heap) for Dijkstra and a 2D visited array.
Complexity:
Time: O(MN \log(MN)) due to priority queue operations.
Space: O(MN) to store the distances/visited status.

Implementation

Wrap Up

Advanced Topics

Memory Optimization: In competitive programming, if M \times N is very large, using a 1D array for minTime (mapping (r, c) to r * n + c) can save overhead.
Wait Logic Generalization: This problem is a variation of finding paths in a graph with time-dependent edge weights. If "waiting in place" was allowed, the parity logic would be unnecessary.
Parallelization: While Dijkstra is inherently sequential, for extremely large grids, the search space can be partitioned or solved using A* with a Manhattan distance heuristic to guide the search towards the bottom-right faster.