Shortest Path in a Grid with Obstacles Elimination

class PathState implements Comparable<PathState> {
    public int costEstimate, moveCount, posX, posY, remainingK;
    private int[] destination;

    public PathState(int moveCount, int posX, int posY, int remainingK, int[] destination) {
        this.moveCount = moveCount;
        this.posX = posX;
        this.posY = posY;
        this.remainingK = remainingK;

        this.destination = destination;
        int heuristic = Math.abs(destination[0] - posX) + Math.abs(destination[1] - posY);
        this.costEstimate = heuristic + moveCount;
    }

    @Override
    public int hashCode() {
        return (this.posX + 1) * (this.posY + 1) * this.remainingK;
    }

    @Override
    public int compareTo(Object obj) {
        PathState other = (PathState) obj;
        return this.costEstimate - other.costEstimate;
    }

    @Override
    public boolean equals(Object obj) {
        if (!(obj instanceof PathState)) return false;
        PathState other = (PathState) obj;
        return this.posX == other.posX && this.posY == other.posY && this.remainingK == other.remainingK;
    }

    @Override
    public String toString() {
        return String.format("(%d %d %d %d %d)",
                this.costEstimate, this.moveCount, this.posX, this.posY, this.remainingK);
    }
}

class PathFinder {

    public int findShortestPath(int[][] matrix, int k) {
        int rowTotal = matrix.length, columnTotal = matrix[0].length;
        int[] targetPoint = {rowTotal - 1, columnTotal - 1};

        PriorityQueue<PathState> priorityQueue = new PriorityQueue<>();
        HashSet<PathState> visited = new HashSet<>();

        PathState initialState = new PathState(0, 0, 0, k, targetPoint);
        priorityQueue.offer(initialState);
        visited.add(initialState);

        while (!priorityQueue.isEmpty()) {
            PathState currentState = priorityQueue.poll();
            int minPossibleCost = currentState.costEstimate - currentState.moveCount;
            if (minPossibleCost <= currentState.remainingK) {
                return currentState.costEstimate;
            }

            int[] moves = {currentState.posX, currentState.posY + 1, currentState.posX + 1, currentState.posY,
                currentState.posX, currentState.posY - 1, currentState.posX - 1, currentState.posY};

            for (int idx = 0; idx < moves.length; idx += 2) {
                int newRow = moves[idx], newCol = moves[idx + 1];

                if (newRow < 0 || newRow >= rowTotal || newCol < 0 || newCol >= columnTotal) continue;

                int nextK = currentState.remainingK - matrix[newRow][newCol];
                PathState nextState = new PathState(currentState.moveCount + 1, newRow, newCol, nextK, targetPoint);

                if (nextK >= 0 && visited.add(nextState)) {
                    priorityQueue.offer(nextState);
                }
            }
        }

        return -1;
    }
}

The provided Java code addresses the problem of finding the shortest path in a grid with possible obstacles that can be removed, up to a specified limit (k times). This challenge is tackled using a search algorithm enhanced with both pathfinding and optimization techniques. Below is a step-wise explanation of how the code works:

1. Define a PathState class to represent each state in the path:

   • PathState includes position coordinates (posX, posY), the number of moves taken (moveCount), the remaining obstacle removals (remainingK), and a heuristic costEstimate based on the estimated path cost from the current position to the destination.
   • This class also implements the Comparable interface to facilitate sorting by costEstimate in a priority queue, enabling efficient retrieval of the lowest-cost option.

2. Implement the PathFinder class with a method findShortestPath which utilizes the PathState:

   • A PriorityQueue is used to process paths in order of their estimated cost from start to the destination (costEstimate), efficiently guiding the search towards promising paths.
   • A HashSet records visited states to avoid redundant processing.
   • The method starts with initializing the path finding process from the top-left corner of the grid (0, 0) using the initial available obstacle removals (k).

3. Employ a heuristic and greedy strategy to manage the exploration:

   • As states are evaluated, the algorithm checks if the potential moves from the current state exceed the remaining allowable obstacle removals.
   • It iterates over possible moves (right, down, left, up), assessing the feasibility of each move regarding grid boundaries and updates the state accordingly.
   • The priority queue ensures states with better heuristic estimates (costEstimate) are processed first, guiding the search towards the target with a more directed approach and potentially reducing the number of states explored.

4. Exit conditions and result handling:

   • If the minimum possible path cost (costEstimate - moveCount) derived during processing of a state is less than or equal to the remainingK, the path's cost is immediately returned.
   • If the queue is exhausted without finding a viable path adhering to the constraints, the method returns -1, indicating no valid paths exist within the provided constraints on obstacle removals.

This approach efficiently balances exploration and exploitation using heuristics and greedy methods, making it suited for scenarios where prioritized exploration of paths in complex spaces with constraints is required.

class Solution:
    def minimumPath(self, grid: List[List[int]], tolerance: int) -> int:

        max_row, max_col = len(grid), len(grid[0])
        destination = (max_row - 1, max_col - 1)

        def distance_to_end(r, c):
            return destination[0] - r + destination[1] - c

        # (current_row, current_col, remaining_tolerance)
        init_state = (0, 0, tolerance)

        # (heuristic, path_length, current_state)
        priority_queue = [(distance_to_end(0, 0), 0, init_state)]
        visited = set([init_state])

        while priority_queue:
            heur, path_len, (curr_row, curr_col, remaining_tolerance) = heapq.heappop(priority_queue)

            minimal_remaining = heur - path_len
            if minimal_remaining <= remaining_tolerance:
                return heur

            # check neighboring cells
            for new_r, new_c in [(curr_row, curr_col + 1), (curr_row + 1, curr_col), (curr_row, curr_col - 1), (curr_row - 1, curr_col)]:
                # remain within grid bounds
                if (0 <= new_r < max_row) and (0 <= new_c < max_col):
                    remaining = remaining_tolerance - grid[new_r][new_c]
                    new_state = (new_r, new_c, remaining)

                    if remaining >= 0 and new_state not in visited:
                        visited.add(new_state)
                        new_heur = distance_to_end(new_r, new_c) + path_len + 1
                        heapq.heappush(priority_queue, (new_heur, path_len + 1, new_state))

        # no path found
        return -1

In the "Shortest Path in a Grid with Obstacles Elimination" problem, the Python solution incorporates the A* search algorithm, using a priority queue to navigate through a grid. The approach relies on a heuristic function to estimate distances and prioritize exploration, and it manages obstacles by accounting for a 'tolerance' level that determines the number of obstacles you can eliminate. Here's a breakdown of the solution's methodology:

• Define the structure of the grid and the tolerance for obstacle elimination. Capture the dimensions of the grid and determine the target destination located at the bottom-right corner of the grid.
• Handle the initial state with the starting point at the top-left corner of the grid and the full tolerance available.
• Utilize a priority queue to keep priority states based on estimated cost from the current cell to the destination through a heuristic function (Manhattan distance) and the path length covered up to that point.
• Implement the search loop, which continues to evaluate adjacent cells until it either finds the destination or depletes all possibilities:
  • For each cell, calculate the heuristic cost, combine it with the traveled path length, and adjust the tolerance based on whether the current cell contains an obstacle.
  • Push neighboring cells into the priority queue if moving into them respects grid boundaries and doesn't exceed obstruction tolerance; mark these states as visited to avoid redundant work.
• The solution evaluates whether the sum of the heuristic and covered path length, adjusted for the minimal expected remaining path under current tolerance, meets the condition to declare the cell part of a valid path towards the destination.
• Return the total minimized path length from the start to the destination if such a path exists under given constraints; otherwise, return -1 indicating no such path can be achieved within the allowed tolerance.

This approach ensures that you maneuver through the grid efficiently while making strategic decisions based on obstacle management and distance remaining, leveraging the combined strategies of breadth-first traversal and heuristic evaluations. With proper handling of edge cases and efficient queue management, this algorithm offers an optimal solution for the grid navigation problem with obstacles.