Minimum Time to Visit a Cell In a Grid

class Solution {
public:
    int computeMinTime(vector<vector<int>>& matrix) {
        // Ensure possibility of immediate move
        if (matrix[0][1] > 1 && matrix[1][0] > 1) {
            return -1;
        }
    
        int numRows = matrix.size(), numCols = matrix[0].size();
        // Define potential move directions
        vector<vector<int>> moves = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        vector<vector<bool>> seen(numRows, vector<bool>(numCols, false));
    
        // Min-heap to consider least time first
        priority_queue<vector<int>, vector<vector<int>>, greater<>> minHeap;
        minHeap.push({matrix[0][0], 0, 0});
    
        while (!minHeap.empty()) {
            auto current = minHeap.top();
            minHeap.pop();
            int currentTime = current[0], currentRow = current[1], currentCol = current[2];
    
            // Destination check
            if (currentRow == numRows - 1 && currentCol == numCols - 1) {
                return currentTime;
            }
    
            // Skip if already checked
            if (seen[currentRow][currentCol]) continue;
            seen[currentRow][currentCol] = true;
    
            for (auto& move : moves) {
                int newRow = currentRow + move[0], newCol = currentCol + move[1];
                if (!isInBounds(seen, newRow, newCol)) continue;
    
                int nextTime = max(matrix[newRow][newCol], currentTime + 1);
                // Directly go to next cell if time matches even seconds after wait
                int requiredTime = ((nextTime - currentTime) % 2 == 0) ? 1 : 0;
                int updatedTime = max(matrix[newRow][newCol] + requiredTime, currentTime + 1);
                minHeap.push({updatedTime, newRow, newCol});
            }
        }
        return -1;
    }
    
private:
    // Validate that the cell is within bounds and unseen
    bool isInBounds(vector<vector<bool>>& seen, int r, int c) {
        return r >= 0 && c >= 0 && r < seen.size() && c < seen[0].size() && !seen[r][c];
    }
};

This C++ solution addresses the problem of finding the minimum time to visit all cells in a given grid from a starting point (top-left corner) to the destination point (bottom-right corner). It utilizes a priority queue to execute a variation of Dijkstra's algorithm optimized for grid traversal. Below are the key components of the implementation:

• Priority Queue: This Min-heap structure helps prioritize moves based on the least total time incurred so far to reach a cell.

• Movement and Boundary Check: The code defines possible moves as vectors (up, down, left, right). For each movement, it calculates the potential new position and checks if this position is within the grid boundaries and has not been visited yet.

• Time Calculation: The solution manually computes the time to visit each cell, considering different conditions like even/odd time intervals and the necessity to wait for the optimum time to enter a future cell, based on its value.

• Seen Matrix: A 2D boolean matrix named "seen" tracks cells that have already been expanded in the search process, avoiding the re-processing of cells.

• Early Exit Conditions: The function contains mechanism to immediately terminate if it reaches the destination cell or if it detects that an initial move is impossible based on the values of the neighboring cells.

• Helper Function: A helper function isInBounds ensures that the proposed move stays within grid limits and that the destination cell has not been visited.

The function returns the minimum time required to traverse from the top-left to the bottom-right of the matrix, or -1 if such a traversal is impossible under the given conditions. This efficient approach ensures that each cell is processed at most once and in the best order relative to its potential total time cost, adhering to the principles of cost-based pathfinding in grid environments.

class Solution {
    
    public int findMinTime(int[][] maze) {
        // Check initial conditions for grid
        if (maze[0][1] > 1 && maze[1][0] > 1) {
            return -1;
        }
    
        int totalRows = maze.length, totalCols = maze[0].length;
        // Directions for moving
        int[][] moves = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
        boolean[][] seen = new boolean[totalRows][totalCols];
    
        // Queue for BFS with cells sorted by time
        PriorityQueue<int[]> minHeap = new PriorityQueue<>((item1, item2) ->
            Integer.compare(item1[0], item2[0])
        );
        minHeap.add(new int[] { maze[0][0], 0, 0 });
    
        while (!minHeap.isEmpty()) {
            int[] current = minHeap.poll();
            int currentTime = current[0], r = current[1], c = current[2];
    
            // Check if target cell is reached
            if (r == totalRows - 1 && c == totalCols - 1) {
                return currentTime;
            }
    
            // Continue if current cell is already processed
            if (seen[r][c]) {
                continue;
            }
            seen[r][c] = true;
    
            // Explore all possible moves
            for (int[] move : moves) {
                int newRow = r + move[0], newCol = c + move[1];
                if (!isWithinBounds(seen, newRow, newCol)) {
                    continue;
                }
    
                // Compute time to enter the neighbors
                int requiredWait = ((maze[newRow][newCol] - currentTime) % 2 == 0) ? 1 : 0;
                int newTime = Math.max(
                    maze[newRow][newCol] + requiredWait,
                    currentTime + 1
                );
                minHeap.add(new int[] { newTime, newRow, newCol });
            }
        }
        return -1;
    }
    
    // Validate if the cell position is within valid range and not visited
    private boolean isWithinBounds(boolean[][] seenState, int r, int c) {
        return (
            r >= 0 &&
            c >= 0 &&
            r < seenState.length &&
            c < seenState[0].length &&
            !seenState[r][c]
        );
    }
}

The solution provides an implementation in Java to find the minimum time required to visit a specific cell in a grid, given certain movement constraints and time costs associated with each cell in the grid. The algorithm applies a Breadth-First Search (BFS) approach utilizing a priority queue to ensure that the cell with the least cumulative time is processed first.

Key components of the solution:

• The algorithm begins by checking the initial conditions of the neighboring cells to the start position. If the initial conditions are not feasible (i.e., both the directly right and directly below cells of the starting position have a value greater than 1), the function returns -1, indicating that it's impossible to proceed.

• A 2D array seen keeps track of the cells that have been visited to avoid processing a cell more than once.

• Movement directions are defined as an array of integer arrays, representing possible moves (right, left, down, up).

• The algorithm uses a priority queue (min-heap) to store cells in the order of their cumulative time to reach them. Initially, the starting cell (0, 0) with its time cost is added to the queue.

• Inside the main while loop, cells are dequeued based on their cumulative time. For each cell, the algorithm checks:

  • If the target cell (bottom-right corner) is reached, which if true, returns the current cumulated time as the result.
  • If the cell has been visited before, it skips processing to avoid redundant calculations.

• For each unvisited cell processed, the algorithm calculates potential new positions based on defined moves, and verifies if they are within grid bounds and not yet visited.

• For every valid move, it computes the waiting time required before the move can be made so that the time conditions are met. This waiting time is the time difference needed to make the current time even or odd as required by the destination cell's time.

• It calculates the new time incorporating the required waiting time and adds the new position along with its computed time back into the queue.

• In the event the queue is exhausted without reaching the target cell, the function returns -1 to indicate no possible path exists under the given constraints.

Conclusion:

The implementation effectively leverages BFS combined with a priority based processing of cells to efficiently determine the minimum time needed to navigate through a grid with specific time-based conditions imposed on each move. This approach ensures optimal movement through the grid by always choosing the next cell based on the minimum time required to reach it.

class Solution:
    def findMinTime(self, matrix: List[List[int]]) -> int:
        if matrix[0][1] > 1 and matrix[1][0] > 1:
            return -1
    
        numRows, numCols = len(matrix), len(matrix[0])
        moves = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        explored = set()
    
        priorityQueue = [(matrix[0][0], 0, 0)]
    
        while priorityQueue:
            t, r, c = heapq.heappop(priorityQueue)
    
            if (r, c) == (numRows - 1, numCols - 1):
                return t
    
            if (r, c) in explored:
                continue
            explored.add((r, c))
    
            for moveR, moveC in moves:
                newR, newC = r + moveR, c + moveC
    
                if not self.isValid(explored, newR, newC, numRows, numCols):
                    continue
    
                addedTime = 1 if (matrix[newR][newC] - t) % 2 == 0 else 0
                newTime = max(matrix[newR][newC] + addedTime, t + 1)
                heapq.heappush(priorityQueue, (newTime, newR, newC))
    
        return -1
    
    def isValid(self, explored, row, col, numRows, numCols):
        return 0 <= row < numRows and 0 <= col < numCols and (row, col) not in explored

The provided Python code aims to find the minimum time required to visit a cell in a grid, following specific rules for movement and time increment. Analyzing the findMinTime function within the Solution class, detailed steps and functionalities can be observed:

1. Start by checking if both cells adjacent to the starting point (top-left corner) are accessible; return -1 if not.
2. Initialize the number of rows and columns derived from the matrix dimensions.
3. Define possible moves within the grid: up, down, left, and right.
4. Use a set to keep track of explored positions to avoid revisiting them.
5. Implement a priority queue initialized with the starting point. The heap data structure ensures that cells with the smallest recorded time are processed first.
6. Enter a loop that continues until the priority queue is empty:
   • Extract the current cell along with its visit time.
   • Check if the destination (bottom-right corner) has been reached; if so, return the current time.
   • If the current cell has already been explored, simply continue to the next iteration.
   • Mark the current cell as explored.
   • For each possible move from the current position, calculate the new position.
   • Use the helper function isValid to check if the new position can be visited: it must be within bounds and not already explored.
   • For cells that pass the validity check, compute the new potential visit time, considering even or odd adjustments.
   • Add the valid new positions along with their computed times to the priority queue.
7. If no valid path is found and the queue empties, return -1.

Additional helper function isValid utilized for checking whether a given position in the grid is valid, ensures that:

• The row and column indices are within the boundaries of the grid.
• The specific cell has not been previously explored.

This function enables the solution class to avoid redundant computations and keep track of its progress efficiently by verifying and processing cells whose visit ensures minimal traversal time within the constraints set by the matrix's format.