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Shortest Path in a Grid with Obstacles Elimination

Hard

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Topics:

Graphs

You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.

Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

Example 1:

Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10.
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2:

Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
Output: -1
Explanation: We need to eliminate at least two obstacles to find such a walk.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 40
1 <= k <= m * n
grid[i][j] is either 0 or 1.
grid[0][0] == grid[m - 1][n - 1] == 0

Shortest Path in a Grid with Obstacles Elimination

Hard

4 views

Topics:

Graphs

You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.

Example 1:

Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10.
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2:

Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
Output: -1
Explanation: We need to eliminate at least two obstacles to find such a walk.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 40
1 <= k <= m * n
grid[i][j] is either 0 or 1.
grid[0][0] == grid[m - 1][n - 1] == 0

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 are the constraints on these dimensions?
What are the possible values for grid cells (e.g., only 0 and 1)? What do 0 and 1 represent, specifically?
What are the constraints on the value of `k` (the maximum number of obstacles you can eliminate)? Is `k` always non-negative?
If there is no path from the top-left to the bottom-right, what should the function return?
Is the starting cell (0, 0) and the ending cell (m-1, n-1) guaranteed to be valid (within the grid bounds) and not an obstacle initially?

Brute Force Solution

Approach

Imagine you're navigating a maze and allowed to break down a certain number of walls. The brute force way involves trying every possible path through the maze, deciding at each step whether to move normally or break a wall. We explore every single route to see which one is the shortest.

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

Start at the beginning of the maze.
Consider every possible move you can make: moving up, down, left, or right.
For each move, check if there's a wall blocking your way.
If there is a wall and you still have wall-breaking abilities left, consider breaking the wall and moving forward.
Keep track of how many steps you've taken and how many walls you've broken on each path.
Repeat the process of considering all possible moves from your new location, until you reach the end of the maze or run out of wall-breaking abilities.
If you reach the end, record the length of the path you took.
Compare all the paths that successfully reached the end and choose the shortest one.

Code Implementation

def shortest_path_brute_force(
    grid, allowed_obstacle_eliminations
):
    rows = len(grid)
    cols = len(grid[0])

    if rows == 0 or cols == 0:
        return -1

    shortest_path = float('inf')

    def explore_path(
        row_index,
        col_index,
        current_path_length,
        remaining_obstacle_eliminations,
    ):
        nonlocal shortest_path

        if row_index < 0 or row_index >= rows or \
           col_index < 0 or col_index >= cols:
            return

        if row_index == rows - 1 and col_index == cols - 1:
            shortest_path = min(
                shortest_path, current_path_length
            )
            return

        # If we hit an obstacle
        if grid[row_index][col_index] == 1:

            # Check if we can eliminate it
            if remaining_obstacle_eliminations > 0:
                explore_path(
                    row_index + 1,
                    col_index,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index - 1,
                    col_index,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index,
                    col_index + 1,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index,
                    col_index - 1,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

        # If its not an obstacle then explore adjacent cells
        else:
            explore_path(
                row_index + 1,
                col_index,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index - 1,
                col_index,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index,
                col_index + 1,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index,
                col_index - 1,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

    explore_path(0, 0, 0, allowed_obstacle_eliminations)

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

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(4^(m*n)) – The provided brute force approach explores every possible path in the grid, considering all combinations of movements and wall eliminations. In the worst case, from each cell, there are up to 4 possible moves (up, down, left, right). The grid has m rows and n columns. Therefore, the number of possible paths grows exponentially with the size of the grid. Since each cell can have 4 choices leading to a branching factor of 4, the total number of paths explored is approximately O(4^(m*n)).

Space Complexity

O(M * N * K) – The algorithm uses a queue to store possible paths to explore. In the worst-case scenario, the queue might contain all possible states, where each state is represented by a cell (row, column) and the remaining number of obstacle eliminations (k). The number of cells is M * N, where M is the number of rows and N is the number of columns. For each cell, there can be K possible values for the number of remaining obstacle eliminations. Therefore, the queue can hold up to M * N * K elements. This dominates the space complexity, thus the auxiliary space is O(M * N * K).

Optimal Solution

Approach

The problem asks for the shortest path through a grid, where you can eliminate a certain number of obstacles. We will explore paths, but we'll prioritize paths that are likely to be shorter and keep track of how many obstacles we've eliminated along each path.

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

Start at the beginning of the grid.
Imagine exploring from your current spot in all possible directions (up, down, left, right).
When you hit an empty space, simply move to that space and remember the path you took.
If you encounter an obstacle, check if you have any 'obstacle elimination' chances left.
If you do, use one chance to eliminate the obstacle and move forward. Keep track of how many chances you have left along this path.
If you don't have any chances left, ignore that direction; it's a dead end.
Prioritize exploring paths that are shorter and have more obstacle elimination chances remaining. This helps you find the shortest path faster.
Keep exploring until you reach the end of the grid. If you find multiple paths to the end, pick the shortest one.
If you've explored all possible paths and haven't found a way to the end, it means there's no solution.

Code Implementation

from collections import deque

def shortest_path_obstacles_elimination(grid, allowed_obstacle_eliminations):
    rows = len(grid)
    cols = len(grid[0])
    queue = deque([(0, 0, allowed_obstacle_eliminations, 0)])
    visited = set()
    visited.add((0, 0, allowed_obstacle_eliminations))

    while queue:
        row_index, column_index, remaining_eliminations, path_length = queue.popleft()

        if row_index == rows - 1 and column_index == cols - 1:
            return path_length

        # Explore all four directions
        for delta_row, delta_col in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            new_row_index = row_index + delta_row
            new_column_index = column_index + delta_col

            if 0 <= new_row_index < rows and 0 <= new_column_index < cols:
                # If empty cell, move
                if grid[new_row_index][new_column_index] == 0:
                    if (new_row_index, new_column_index, remaining_eliminations) not in visited:
                        queue.append((new_row_index, new_column_index, remaining_eliminations, path_length + 1))
                        visited.add((new_row_index, new_column_index, remaining_eliminations))

                else:
                    # Check if we can eliminate
                    if remaining_eliminations > 0:
                        # Reduce obstacle eliminations if cell is blocked
                        if (new_row_index, new_column_index, remaining_eliminations - 1) not in visited:
                            queue.append((new_row_index, new_column_index, remaining_eliminations - 1, path_length + 1))
                            visited.add((new_row_index, new_column_index, remaining_eliminations - 1))

    return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m*n*k) – The algorithm explores the grid using a modified breadth-first search (BFS) where m is the number of rows, and n is the number of columns of the grid. Each cell (r, c) can be visited with different remaining obstacle eliminations (from 0 to k). Thus, in the worst-case scenario, we could potentially visit each cell multiple times (up to k+1 times) with different values for remaining obstacle elimination opportunities. Therefore, the time complexity is proportional to the number of cells in the grid multiplied by the maximum number of obstacle eliminations allowed, resulting in O(m*n*k) where k is the maximum obstacle eliminations allowed.

Space Complexity

O(M * N * K) – The algorithm uses a queue or similar data structure to store the paths being explored. In the worst case, we might enqueue every cell in the grid (M rows and N columns) for each possible value of obstacle eliminations K, resulting in M * N * K states being stored in the queue. Additionally, we likely need a visited set to track cells and obstacle eliminations already explored, which would also take up to O(M * N * K) space. Therefore, the auxiliary space complexity is O(M * N * K), where M and N represent the dimensions of the grid, and K is the maximum number of obstacle eliminations allowed.

Edge Cases

Case	How to Handle
Null or empty grid	Return -1 immediately as there is no path.
Invalid k (negative k or k less than 0)	Return -1 since a negative number of eliminations is not valid.
Start equals end (already at destination)	Return 0 as no moves are required.
No obstacles (all grid cells are 0)	Return the Manhattan distance from start to end, which is the shortest path.
k is greater than or equal to the number of obstacles on any possible path	Return the Manhattan distance as we can eliminate all obstacles on the direct path.
No valid path exists even with k obstacle eliminations	Return -1 when the queue is exhausted without reaching the target.
Large grid dimensions causing potential memory issues (BFS queue size)	Implement a more memory-efficient data structure or optimize BFS to prune unnecessary states.
Integer overflow when calculating distances in very large grids	Use a data type with a larger range (e.g., long) to store distances and obstacle eliminations.

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 are the constraints on these dimensions?
What are the possible values for grid cells (e.g., only 0 and 1)? What do 0 and 1 represent, specifically?
What are the constraints on the value of `k` (the maximum number of obstacles you can eliminate)? Is `k` always non-negative?
If there is no path from the top-left to the bottom-right, what should the function return?
Is the starting cell (0, 0) and the ending cell (m-1, n-1) guaranteed to be valid (within the grid bounds) and not an obstacle initially?

Brute Force Solution

Approach

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

Start at the beginning of the maze.
Consider every possible move you can make: moving up, down, left, or right.
For each move, check if there's a wall blocking your way.
If there is a wall and you still have wall-breaking abilities left, consider breaking the wall and moving forward.
Keep track of how many steps you've taken and how many walls you've broken on each path.
Repeat the process of considering all possible moves from your new location, until you reach the end of the maze or run out of wall-breaking abilities.
If you reach the end, record the length of the path you took.
Compare all the paths that successfully reached the end and choose the shortest one.

Code Implementation

def shortest_path_brute_force(
    grid, allowed_obstacle_eliminations
):
    rows = len(grid)
    cols = len(grid[0])

    if rows == 0 or cols == 0:
        return -1

    shortest_path = float('inf')

    def explore_path(
        row_index,
        col_index,
        current_path_length,
        remaining_obstacle_eliminations,
    ):
        nonlocal shortest_path

        if row_index < 0 or row_index >= rows or \
           col_index < 0 or col_index >= cols:
            return

        if row_index == rows - 1 and col_index == cols - 1:
            shortest_path = min(
                shortest_path, current_path_length
            )
            return

        # If we hit an obstacle
        if grid[row_index][col_index] == 1:

            # Check if we can eliminate it
            if remaining_obstacle_eliminations > 0:
                explore_path(
                    row_index + 1,
                    col_index,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index - 1,
                    col_index,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index,
                    col_index + 1,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

                explore_path(
                    row_index,
                    col_index - 1,
                    current_path_length + 1,
                    remaining_obstacle_eliminations - 1,
                )

        # If its not an obstacle then explore adjacent cells
        else:
            explore_path(
                row_index + 1,
                col_index,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index - 1,
                col_index,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index,
                col_index + 1,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

            explore_path(
                row_index,
                col_index - 1,
                current_path_length + 1,
                remaining_obstacle_eliminations,
            )

    explore_path(0, 0, 0, allowed_obstacle_eliminations)

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

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the beginning of the grid.
Imagine exploring from your current spot in all possible directions (up, down, left, right).
When you hit an empty space, simply move to that space and remember the path you took.
If you encounter an obstacle, check if you have any 'obstacle elimination' chances left.
If you do, use one chance to eliminate the obstacle and move forward. Keep track of how many chances you have left along this path.
If you don't have any chances left, ignore that direction; it's a dead end.
Prioritize exploring paths that are shorter and have more obstacle elimination chances remaining. This helps you find the shortest path faster.
Keep exploring until you reach the end of the grid. If you find multiple paths to the end, pick the shortest one.
If you've explored all possible paths and haven't found a way to the end, it means there's no solution.

Code Implementation

from collections import deque

def shortest_path_obstacles_elimination(grid, allowed_obstacle_eliminations):
    rows = len(grid)
    cols = len(grid[0])
    queue = deque([(0, 0, allowed_obstacle_eliminations, 0)])
    visited = set()
    visited.add((0, 0, allowed_obstacle_eliminations))

    while queue:
        row_index, column_index, remaining_eliminations, path_length = queue.popleft()

        if row_index == rows - 1 and column_index == cols - 1:
            return path_length

        # Explore all four directions
        for delta_row, delta_col in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            new_row_index = row_index + delta_row
            new_column_index = column_index + delta_col

            if 0 <= new_row_index < rows and 0 <= new_column_index < cols:
                # If empty cell, move
                if grid[new_row_index][new_column_index] == 0:
                    if (new_row_index, new_column_index, remaining_eliminations) not in visited:
                        queue.append((new_row_index, new_column_index, remaining_eliminations, path_length + 1))
                        visited.add((new_row_index, new_column_index, remaining_eliminations))

                else:
                    # Check if we can eliminate
                    if remaining_eliminations > 0:
                        # Reduce obstacle eliminations if cell is blocked
                        if (new_row_index, new_column_index, remaining_eliminations - 1) not in visited:
                            queue.append((new_row_index, new_column_index, remaining_eliminations - 1, path_length + 1))
                            visited.add((new_row_index, new_column_index, remaining_eliminations - 1))

    return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty grid	Return -1 immediately as there is no path.
Invalid k (negative k or k less than 0)	Return -1 since a negative number of eliminations is not valid.
Start equals end (already at destination)	Return 0 as no moves are required.
No obstacles (all grid cells are 0)	Return the Manhattan distance from start to end, which is the shortest path.
k is greater than or equal to the number of obstacles on any possible path	Return the Manhattan distance as we can eliminate all obstacles on the direct path.
No valid path exists even with k obstacle eliminations	Return -1 when the queue is exhausted without reaching the target.
Large grid dimensions causing potential memory issues (BFS queue size)	Implement a more memory-efficient data structure or optimize BFS to prune unnecessary states.
Integer overflow when calculating distances in very large grids	Use a data type with a larger range (e.g., long) to store distances and obstacle eliminations.