Minimum Cost to Make at Least One Valid Path in a Grid

Hard

Google

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Topics:

GraphsDynamic Programming

You are given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1]).
2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1]).
3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j]).
4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j]).

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0

Can you provide an algorithm to solve this problem efficiently?

Minimum Cost to Make at Least One Valid Path in a Grid

Hard

Google

1 view

Topics:

GraphsDynamic Programming

You are given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1]).
2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1]).
3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j]).
4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j]).

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0

Can you provide an algorithm to solve this problem efficiently?

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 dimensions of the grid (number of rows and columns), and what is the maximum size they can be?
What is the range of possible values for the cost of changing a direction (i.e., the value in the grid cell)? Can it be zero or negative?
If it's impossible to find a valid path from the top-left to the bottom-right, what should I return?
Are the grid dimensions guaranteed to be non-zero, i.e., will the number of rows and columns always be at least 1?
By 'valid path', do you mean a path from the top-left cell (0, 0) to the bottom-right cell (m-1, n-1) where 'm' is the number of rows and 'n' is the number of columns, following the directions indicated in each cell without incurring any cost, or do we only need at least one cell that follows the correct direction?

Brute Force Solution

Approach

The brute force approach to this grid problem means we're going to explore every single possible way to make a path from the start to the finish. We'll try changing the direction of arrows and calculate the cost of each of these attempted paths, keeping track of whether or not the paths are valid.

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

Start at the beginning of the grid.
Consider all possible routes you can take by following the arrows as they are or changing their direction.
For each time you change the direction of an arrow, increase your cost by one.
Continue exploring each route until you reach the end of the grid or you get stuck.
If you reach the end of the grid, check if it's a valid path. If so, remember the total cost.
Repeat this process by starting again at the beginning and exploring every other possible combination of arrow direction changes.
After trying every possible path, compare the costs of all the valid paths you found.
Choose the path with the lowest cost. That's the minimum cost to make at least one valid path.

Code Implementation

def minimum_cost_valid_path_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])

    minimum_cost = float('inf')

    def explore_paths(current_row, current_col, current_cost, current_path):
        nonlocal minimum_cost

        # Base case: Reached the destination
        if current_row == rows - 1 and current_col == cols - 1:
            minimum_cost = min(minimum_cost, current_cost)
            return

        # Base case: Out of bounds
        if current_row < 0 or current_row >= rows or current_col < 0 or current_col >= cols:
            return

        # Avoid cycles by tracking visited cells
        if (current_row, current_col) in current_path:
            return

        current_path.add((current_row, current_col))
        
        # Explore all possible directions (right, left, down, up) and costs
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        for index, (row_change, col_change) in enumerate(directions):
            new_row = current_row + row_change
            new_col = current_col + col_change

            cost_increment = 0
            # If the current direction doesn't match the grid's direction, increment the cost.
            if grid[current_row][current_col] != index + 1:
                cost_increment = 1

            explore_paths(new_row, new_col, current_cost + cost_increment, current_path.copy())

    explore_paths(0, 0, 0, set())

    if minimum_cost == float('inf'):
        return -1
    else:
        return minimum_cost

Big(O) Analysis

Time Complexity

O(4^(m*n)) – The brute force approach explores every possible path in the m x n grid by potentially changing the direction of each arrow. For each cell, we can choose to either follow the existing direction or change it to one of the other three directions. This means for each cell, there are 4 possible choices. Since there are m*n cells, the total number of possible paths explored is on the order of 4 raised to the power of (m*n). Thus, the time complexity is O(4^(m*n)).

Space Complexity

O(2^N) – The brute force approach explores every possible path in the grid by changing arrow directions. Since each cell has a fixed number of direction change options, and we consider all combinations, the space complexity becomes exponential, as a recursion tree is inherently built. This recursion tree's depth could potentially reach a level where we are exploring different direction changes and each level will represent a choice in the direction change. Let's assume N is the total number of cells in the grid; in the worst-case scenario, the recursion depth could be proportional to N, thus there could be 2^N possible routes stored implicitly by the recursion stack. Thus the auxiliary space used by the recursion stack is O(2^N).

Optimal Solution

Approach

This problem asks for the least expensive way to make sure there's at least one path from the start to the end of a grid, following arrow directions. The optimal solution uses a clever twist on a well-known pathfinding algorithm to prioritize paths that match the given directions.

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

Imagine the grid as a map with roads and costs to change directions.
Start at the beginning and explore the grid, always trying to follow the arrows already present.
When you have to change the direction of an arrow, think of it as adding a cost to your path.
Use a special technique that explores all possible routes simultaneously, always favoring those that have the lowest cost so far.
As you explore, keep track of the lowest cost to reach each location on the grid.
The first time you reach the end of the grid, you've found the minimum cost to make at least one valid path.

Code Implementation

import heapq

def minimum_cost_to_make_at_least_one_valid_path_in_a_grid(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    # Define directions: right, left, down, up
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    # Initialize cost matrix with infinity
    cost = [[float('inf')] * number_of_columns for _ in range(number_of_rows)]
    cost[0][0] = 0

    priority_queue = [(0, 0, 0)]  # (cost, row, column)

    while priority_queue:
        current_cost, current_row, current_column = heapq.heappop(priority_queue)

        if current_cost > cost[current_row][current_column]:
            continue

        if current_row == number_of_rows - 1 and current_column == number_of_columns - 1:
            return current_cost

        for direction_index, (row_change, column_change) in enumerate(directions):
            new_row = current_row + row_change
            new_column = current_column + column_change

            if 0 <= new_row < number_of_rows and 0 <= new_column < number_of_columns:
                # Calculate new cost: 0 if direction matches, 1 otherwise
                new_cost = current_cost + (grid[current_row][current_column] != direction_index + 1)

                # Update cost and push to queue if a better path is found
                if new_cost < cost[new_row][new_column]:
                    cost[new_row][new_column] = new_cost
                    heapq.heappush(priority_queue, (new_cost, new_row, new_column))

    return -1 # No path found

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm utilizes a 0-1 breadth-first search (BFS) or Dijkstra-like approach to explore the grid. In the worst-case scenario, every cell in the m x n grid could be visited and enqueued/dequeued. Each cell is visited at most once because the algorithm keeps track of the minimum cost to reach each cell. Therefore, the time complexity is proportional to the number of cells in the grid, resulting in O(m*n) time complexity where m is the number of rows and n is the number of columns. Each enqueue/dequeue operation is O(1).

Space Complexity

O(M*N) – The algorithm uses a priority queue (or similar data structure for the 'special technique') to explore the grid, storing nodes to visit and their associated costs. In the worst case, this queue could contain all the cells of the grid. Also, the algorithm keeps track of the lowest cost to reach each location on the grid, implying a cost matrix (or equivalent data structure) of the same size as the grid. Therefore, the auxiliary space is proportional to the number of cells in the grid, M*N, where M is the number of rows and N is the number of columns. This gives a space complexity of O(M*N).

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 if the grid is null or empty, as there is no path to modify.
1x1 grid	Return 0, since the path is already valid (or trivially so).
Grid with only one row or one column	Iterate through the row/column and count the number of incorrect directions, summing the cost.
Large grid (potential memory/time issues)	The A* search or Dijkstra's algorithm should scale reasonably well, but consider memory usage for the priority queue and distance matrix.
All cells have the same direction	The cost will be the number of steps to reach the end, minus one if the initial direction is correct.
No path exists even after changing all directions	This is impossible, as changing all signs allows you to zig-zag to the end, the algorithm should still correctly produce the minimum number of changes needed for a path to exist, which will be less than the number of cells.
Integer overflow when calculating cost for large grids	Use a data type that can accommodate large numbers, such as long, to store the cost.
Grid contains invalid direction numbers (e.g., outside 1-4)	Treat invalid direction numbers the same way we treat mismatched directions during cost calculations.