Minimum Path Cost in a Hidden Grid

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GraphsDynamic Programming

You are given a hidden grid of size m x n. You have an initial position of (row, col) which is unknown. You are also given a list of moves where each moves[i] = 'L', 'R', 'U', or 'D' represents the direction to move in. All moves are valid on the grid.

You should trace out the path of the moves on the grid. You are not allowed to go out of bounds of the grid.

You are also given a cost matrix where cost[i][j] is the cost of moving to the cell (i, j). You should calculate the minimum cost of tracing out the path of the moves on the grid, starting from the initial position (row, col).

Return the minimum cost to traverse the path. If it is impossible to traverse the path, return -1.

Note: The initial position of (row, col) is hidden. You will only know the size of the grid m x n.

Example 1:

Input: moves = ["U","D","D"], cost = [[1,1,5],[1,2,4],[1,3,6],[1,4,7],[2,2,9]]
Output: 6
Explanation: The starting position can be (0, 0), (0, 1), (1, 0), (1, 1), or (2, 1).
- Starting at (0, 0): Path is (0, 0) -> (0, 0) -> (1, 0) -> (2, 0). Not valid since you go out of bounds.
- Starting at (0, 1): Path is (0, 1) -> (0, 1) -> (1, 1) -> (2, 1). Cost is 1 + 1 + 2 + 3 = 7.
- Starting at (1, 0): Path is (1, 0) -> (0, 0) -> (1, 0) -> (2, 0). Cost is 1 + 1 + 1 + 1 = 4.
- Starting at (1, 1): Path is (1, 1) -> (0, 1) -> (1, 1) -> (2, 1). Cost is 2 + 1 + 2 + 3 = 8.
- Starting at (2, 1): Path is (2, 1) -> (1, 1) -> (2, 1) -> (3, 1). Cost is 3 + 2 + 3 + 4 = 12. Not valid since you go out of bounds.
The minimum cost of the valid paths is 4.

Example 2:

Input: moves = ["L","L","U","U"], cost = [[2,2,4],[3,4,9],[2,3,6]]
Output: -1
Explanation: It is impossible to generate a valid path.

Constraints:

1 <= moves.length <= 1000
cost.length == m
cost[i].length == n
1 <= m, n <= 100
1 <= cost[i][j] <= 100
moves[i] is either 'L', 'R', 'U', or 'D'.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the grid's dimensions, and the range of possible costs for moving in each direction?
If no path exists to the target location, what should I return?
Are there any restrictions on revisiting cells in the grid? Can I revisit them, or must the path be acyclic?
Is the starting location always at the top-left corner (0, 0), and is the target location always at the bottom-right corner (n-1, m-1), where n and m are the dimensions of the grid?
Can you elaborate on how the 'hidden grid' is represented and how I will receive information about the cost of moving in each direction from a given cell?

Brute Force Solution

Approach

Imagine you're exploring a maze where you can only move in certain directions. The brute force way is to try every possible path from the start to the end. We'll explore each path fully until we hit a dead end or reach the destination.

Here's how the algorithm would work step-by-step:

Begin at the starting point.
Consider all the possible directions you can move from your current location.
For each direction, take a step and remember the cost of that step.
From this new location, again consider all possible directions to move.
Keep moving and adding the cost until you either reach the destination or hit a dead end (a spot where you cannot move further).
If you reach the destination, save the total cost of that path.
If you hit a dead end, go back to the last place you had a choice of direction and try a different direction.
Repeat this process until you've tried absolutely every single possible path from the start to the end.
Finally, compare the costs of all the paths that reached the destination and pick the path with the lowest cost.

Code Implementation

def find_minimum_path_cost_brute_force(grid, start_row, start_col, end_row, end_col, move_cost):
    minimum_cost = float('inf')

    def explore_path(current_row, current_col, current_cost, visited):
        nonlocal minimum_cost

        # If we reach the destination, update min
        if current_row == end_row and current_col == end_col:
            minimum_cost = min(minimum_cost, current_cost)
            return

        # If we go out of bounds or revisit, stop
        if current_row < 0 or current_row >= len(grid) or \
           current_col < 0 or current_col >= len(grid[0]) or \
           (current_row, current_col) in visited:
            return

        visited.add((current_row, current_col))

        # Check each neighbor
        possible_moves = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for delta_row, delta_col in possible_moves:
            new_row = current_row + delta_row
            new_col = current_col + delta_col

            # Calculate cost of the current move
            current_node_value = grid[current_row][current_col]
            move_cost_value = move_cost[current_node_value][delta_row + 1 if delta_row != 0 else 0][delta_col + 1 if delta_col != 0 else 0]

            explore_path(new_row, new_col, current_cost + move_cost_value, visited.copy())

    explore_path(start_row, start_col, 0, set())
    return minimum_cost

Big(O) Analysis

Time Complexity

O(4^N) – The provided solution uses a brute-force approach to explore all possible paths in the grid. In the worst-case scenario, the algorithm could potentially visit every cell multiple times. Each cell has up to four possible directions to move (up, down, left, right). Therefore, the number of possible paths grows exponentially with the maximum path length which is related to the number of cells N in the grid. This leads to a time complexity of approximately O(4^N), where N represents the number of cells in the grid, reflecting the exponential growth of paths explored.

Space Complexity

O(N) – The algorithm uses recursion to explore each possible path. In the worst-case scenario, where the path resembles a straight line or has minimal branching, the recursion depth could reach N, where N represents the number of cells in the longest possible path in the grid (potentially all the cells if the grid is fully connected and the destination is far from the start). Each recursive call adds a new frame to the call stack, consuming memory. Therefore, the auxiliary space used by the recursion stack is proportional to N, giving a space complexity of O(N).

Optimal Solution

Approach

Imagine exploring a maze where you don't know the map beforehand. The best way to find the cheapest path is to gradually explore, always choosing the direction that seems cheapest so far. This approach guarantees you'll eventually find the overall cheapest route.

Here's how the algorithm would work step-by-step:

Start at the beginning point of the hidden grid.
Explore each possible direction (up, down, left, right) from your current location. Note the cost to move in each direction.
Choose the direction with the lowest cost and move there. Mark this new spot as visited.
Repeat the process: explore the possible directions from this new location, noting the costs. But, ignore paths that lead back to places you've already visited.
Continue moving to the cheapest unvisited location until you reach your destination.
The total cost of your path is the sum of the costs of each move you made along the way. Since you always chose the cheapest available option at each step, this will be the minimum possible cost.

Code Implementation

def find_minimum_path_cost(
        grid_interface, start_node, target_node):

    current_position = start_node
    visited_nodes = {start_node}
    total_cost = 0

    while current_position != target_node:
        possible_moves = []
        
        for direction in ['U', 'D', 'L', 'R']:
            cost = grid_interface.move(direction)
            if cost != -1:
                next_node = grid_interface.get_position()
                if next_node not in visited_nodes:
                    possible_moves.append((cost, direction, next_node))
                grid_interface.move(get_opposite_direction(direction))

        # Need to ensure there are valid moves
        if not possible_moves:
            return -1

        best_move_cost, best_move_direction, next_position = min(possible_moves)

        # Moving to the next node with the lowest cost.
        grid_interface.move(best_move_direction)
        total_cost += best_move_cost
        current_position = next_position
        visited_nodes.add(current_position)

    return total_cost

def get_opposite_direction(direction):
    if direction == 'U':
        return 'D'
    elif direction == 'D':
        return 'U'
    elif direction == 'L':
        return 'R'
    else:
        return 'L'

Big(O) Analysis

Time Complexity

O(R*C*log(R*C)) – The algorithm explores a grid of size R rows and C columns. In the worst case, it may visit every cell in the grid. For each visited cell, we explore its four neighbors (up, down, left, right). The crucial aspect impacting complexity is the use of a priority queue (implicitly or explicitly within the 'choose the direction with the lowest cost' step) to select the next cell to visit. Adding or retrieving the minimum cost cell from the priority queue takes logarithmic time relative to the number of cells currently in the priority queue. In the worst case, the priority queue might contain all the unvisited cells, which can be up to R*C cells. Therefore, visiting R*C cells, each potentially involving a logarithmic priority queue operation of log(R*C), results in a total time complexity of O(R*C*log(R*C)).

Space Complexity

O(N) – The algorithm keeps track of visited locations to avoid cycles. This requires a data structure, such as a set or a boolean grid, to store whether each cell in the grid has been visited. In the worst case, the algorithm might visit every cell in the grid before reaching the destination. If we define N as the number of cells in the grid, this data structure will consume space proportional to N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty cost matrix or move cost matrix	Return -1 to indicate no path exists if either matrix is empty
Start node is invalid (out of bounds of cost matrix)	Return -1 if the start node is outside the bounds of the cost matrix.
Cost matrix with only one row/column (essentially a 1D array)	The algorithm should still correctly traverse and accumulate costs in the single row/column.
Move cost matrix contains negative costs	Dijkstra can handle negative costs if there are no negative cycles, otherwise algorithm needs adaptation or return error if negative cycle is detected
Cost matrix contains very large or very small cost values (potential overflow)	Use a data type large enough to prevent integer overflow, such as long, or consider using a modulo operation if the question specifies constraints.
All possible paths from the start node are blocked (no valid path to the target)	The algorithm should eventually terminate and return -1 when the priority queue is empty, indicating no path was found.
Maximum size cost and move cost matrices where memory becomes a bottleneck	Optimize memory usage by using appropriate data structures and avoiding unnecessary copies or storage; if memory is exceeded, an appropriate error should be returned.
Move cost matrix causes cycles in the hidden grid traversal	Dijkstra algorithm prevents infinite loops and revisiting previously visited nodes; no changes are needed.

Empty cost matrix or move cost matrix

How to Handle:

Return -1 to indicate no path exists if either matrix is empty

Start node is invalid (out of bounds of cost matrix)

How to Handle:

Return -1 if the start node is outside the bounds of the cost matrix.

Cost matrix with only one row/column (essentially a 1D array)

How to Handle:

The algorithm should still correctly traverse and accumulate costs in the single row/column.

Move cost matrix contains negative costs

How to Handle:

Dijkstra can handle negative costs if there are no negative cycles, otherwise algorithm needs adaptation or return error if negative cycle is detected

Cost matrix contains very large or very small cost values (potential overflow)

How to Handle:

Use a data type large enough to prevent integer overflow, such as long, or consider using a modulo operation if the question specifies constraints.

All possible paths from the start node are blocked (no valid path to the target)

How to Handle:

The algorithm should eventually terminate and return -1 when the priority queue is empty, indicating no path was found.

Maximum size cost and move cost matrices where memory becomes a bottleneck

How to Handle:

Optimize memory usage by using appropriate data structures and avoiding unnecessary copies or storage; if memory is exceeded, an appropriate error should be returned.

Move cost matrix causes cycles in the hidden grid traversal

How to Handle:

Dijkstra algorithm prevents infinite loops and revisiting previously visited nodes; no changes are needed.