Problem Description

You are given a 2D integer array grid with two possible values in each cell:

• 0 represents an empty cell you can move through.
• 1 represents an obstacle that can be removed.

The grid has m rows and n columns, and the goal is to move from the top-left corner (0, 0) to the bottom-right corner (m - 1, n - 1). You can move in four directions from an empty cell: up, down, left, or right.

The problem asks you to find the minimum number of obstacles (1s) that need to be removed to enable this path-finding from the start to the end.

Flowchart Walkthrough

Let's analyze the problem from LeetCode 2290, "Minimum Obstacle Removal to Reach Corner," using the algorithm Flowchart. Here's a step-by-step decomposition:

Is it a graph?

• Yes: The grid can be treated as a graph where each cell is a node, and the movement between cells are the edges.

Is it a tree?

• No: Given the grid layout, it's not specifically a tree as it could contain loops and multiple paths between cells.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The grid allows for movements in multiple directions (not necessarily one-directional or acyclic).

Is the problem related to shortest paths?

• Yes: We are asked to find the minimum number of obstacle removals needed to move from the top-left corner to the bottom-right corner, which is essentially finding the shortest path in terms of obstacle removals.

Is the graph weighted?

• Yes: Since the number of obstacles can vary per cell, and removing these obstacles has a 'cost', this effectively makes the graph weighted where the weights represent the cost (number of obstacles) to move through each cell.

Conclusion: According to the flowchart, for a problem dealing with weighted graphs and shortest paths, we should use Dijkstra's Algorithm. However, since weights are limited (0 or 1 in most cases) and the objective is to find the minimum number, the Breadth-First Search (BFS) with a priority queue can be efficient and is appropriate. Thus, BFS would be a more simplified approach considering the discrete and small range of edge weights (0 or 1). This approach uses BFS to explore the least 'costly' paths first, akin to Dijkstra's algorithm in principle, but adapted to the specific characteristics of the problem's grid and obstacle setup.

Intuition

To solve this problem, we use a breadth-first search (BFS) approach. BFS is a graph traversal method that expands and examines all nodes of a level before moving to the next level. This characteristic makes BFS suitable for finding the shortest path on unweighted graphs, or in this case, to find the minimum number of obstacles to remove.

The idea is to traverse the grid using BFS, keeping track of the position (i, j) we're currently in and the count k of obstacles that we have removed to reach this position. Each time we can either move to an adjacent empty cell without incrementing the obstacle removal count or move to an adjacent cell with an obstacle by incrementing the removal count by one.

We use a deque to facilitate the BFS process. When we encounter an empty cell, we add the position to the front of the deque, giving priority to such moves for expansion before those requiring obstacle removal, effectively ensuring the BFS prioritizes paths with fewer obstacles removed.

A visited set vis is used to keep track of the cells we have already processed to avoid re-processing and potentially getting into cycles.

The search continues until we reach the bottom-right corner (m - 1, n - 1), at which point we return the count of removed obstacles k which represents the minimum number of obstacles that need to be removed.

Learn more about Breadth-First Search, Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The provided Python code implements the BFS algorithm using a deque, a double-ended queue that allows the insertion and removal of elements from both the front and the back with constant time complexity O(1).

BFS (Breadth-First Search)

At a high level, the BFS algorithm works by visiting all neighbors of a node before moving on to the neighbors' neighbors. It uses a queue to keep track of which nodes to visit next.

Deque for BFS

In the Python code, the deque is used instead of a regular queue. When the BFS explores a cell with an obstacle (value 1), it appends the new state to the end of the deque. Conversely, when it explores a cell without an obstacle (value 0), it appends the new state to the front of the deque.

Implementation Details

1. Initialize the deque q with a tuple containing the starting position and the initial number of obstacles removed (0, 0, 0).
2. Create a vis set to record visited positions and avoid revisiting them.
3. The dirs tuple is used to calculate the adjacent cells' positions with the help of a helper function like pairwise.
4. Enter a loop that continues until the end condition is met (reaching (m - 1, n - 1)).
5. Use popleft to get the current position and the count k of the obstacles removed.
6. If the target position is reached, return k.
7. If the position has been visited before ((i, j) in vis), just continue to the next iteration.
8. Otherwise, mark the position as visited by adding (i, j) to vis.
9. Iterate over all possible directions, and calculate the adjacent cell positions (x, y).
10. If (x, y) is within the grid bounds and is not an obstacle, append it to the front of the deque with the same number of removed obstacles k.
11. If (x, y) has an obstacle, append it to the end of the deque and increment the count of removed obstacles by 1.

This process prioritizes exploring paths with fewer obstacles, and since BFS is used, it guarantees that the first time we reach the bottom-right corner is via the path that requires the minimum number of obstacles to be removed.

The solution effectively mixes BFS traversal with a priority queue concept by using a deque to manage traversal order, ensuring an efficient search for the minimum obstacle removal path.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have the following grid:

grid = [
  [0, 1, 0],
  [0, 1, 1],
  [0, 0, 0]
]

Here, m = 3 and n = 3. We want to move from (0, 0) to (2, 2) with the minimum number of obstacles removed.

Step-by-Step Process:

1. We start by initializing the deque q with the starting position and the initial number of obstacles removed, which is 0. So q = deque([(0, 0, 0)]).

2. The vis set is initialized to ensure we don't visit the same positions repeatedly. In the beginning, it's empty: vis = set().

3. The BFS begins. We dequeue the first element (i, j, k) = (0, 0, 0), where (i, j) represents the current cell, and k is the number of removed obstacles so far.

4. We have not reached the target, so we check the adjacent cells:

   • Right (0, 1): It has an obstacle. We increment obstacle count and add it to the end of q: q = deque([(0, 1, 1)]).
   • Down (1, 0): No obstacle. We add it to the front of q: q = deque([(1, 0, 0), (0, 1, 1)]).

5. We mark the current cell (0, 0) as visited: vis.add((0, 0)).

6. Now the deque q has cells to process, so we take the one from the front, which is (1, 0, 0).

7. From (1, 0), we again check the adjacent cells:

   • Right (1, 1): It has an obstacle, add it to end: q = deque([(0, 1, 1), (1, 1, 1)]).
   • Down (2, 0): No obstacle, add it to front: q = deque([(2, 0, 0), (0, 1, 1), (1, 1, 1)]).

8. We mark the cell (1, 0) as visited: vis.add((1, 0)).

9. The steps repeat, dequeuing from the front, checking adjacent cells, and enqueueing in the deque according to the rules.

10. Eventually, we dequeue the cell (2, 0, 0). From here, we can go right to (2, 1) with no obstacle and add it to the front: q = deque([(2, 1, 0), (0, 1, 1), (1, 1, 1)]).

11. Again, we mark (2, 0) as visited: vis.add((2, 0)).

12. Dequeuing (2, 1, 0), we can go right to the target (2, 2) with no obstacle. We add it to the front.

13. Mark (2, 1) as visited and check the next cell from q.

14. Finally, we reach (2, 2) which is the target. We have not needed to remove any obstacles, so we return k = 0.

Throughout this process, we have explored paths with the fewest obstacles first, using the deque to effectively prioritize cells without obstacles. By doing this, we've ensured that the first time we reach the end, it is the path with the minimum number of obstacles removed. In this case, no obstacles needed to be removed.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the algorithm can be estimated by considering the number of operations it performs. The algorithm uses a queue that can have at most every cell (i, j) from the grid enqueued, and the grid has m * n cells. Each cell is inserted into the queue at most once, since we're using a set of visited nodes to prevent re-visitation.

The deque operations appendleft and popleft have O(1) time complexity. The for loop executes at most four times for each cell, corresponding to the four possible directions one can move in the grid.

Considering all of this, the overall time complexity is O(m * n), as every cell is considered at most once.

Space Complexity

The space complexity consists of the space needed for the queue and the set of visited cells.

The queue can store up to m * n elements, if all cells are inserted, and the set will at the same time contain a maximum of m * n elements as well to track visited cells. The sum is 2 * m * n but in terms of Big O notation, the constants are ignored.

Thus, the space complexity is O(m * n) for the queue and the visited set combined.

Learn more about how to find time and space complexity quickly using problem constraints.