What is the brute force approach and complexity for Shortest Path in a Grid with Obstacles Elimination?

DFS over simple paths; each path carries how many obstacles it may still remove with time complexity O(3^(m · n)) and space complexity O(m · n).

What is the optimal approach and complexity for Shortest Path in a Grid with Obstacles Elimination?

BFS over states (cell, removals left); skip a state that a better one already covers with optimal time complexity O(m · n · k) and space complexity O(m · n · k).

How do you identify the pattern for Shortest Path in a Grid with Obstacles Elimination?

Look for key signals in the prompt: "remove up to k obstacles" indicates The removals are a budget. A state has to carry how many are left.; "fewest moves" indicates Every move costs one, so BFS gives the fewest moves. The state is bigger than the cell alone.; "up to k removals" indicates More removals left is never worse. That rule lets the search skip weaker states..

Shortest Path in a Grid with Obstacles Elimination

Hard·LeetCode #1293 Pro OnlySlight Variation

LeetCode

Problem Statement

You are given an m × n grid. A 0 is an empty cell and a 1 is an obstacle. You start at the top-left cell and want to reach the bottom-right cell.

Each move goes up, down, left or right into a neighbouring cell. You may remove up to k obstacles on the way: stepping onto an obstacle and removing it is allowed while you still have removals left. The start and end cells are empty.

Return the fewest moves needed to reach the bottom-right cell. Return -1 if no walk does it with at most k removals.

Input

grid, an m × n grid of 0 and 1, and k, the number of obstacles you may remove.

Output

The fewest moves from the top-left to the bottom-right cell, or -1.

Example 1

Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1

Output: 6

Going round the obstacles without removing any takes 10 moves. Removing the obstacle at (3,2) gives a 6-move walk: (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

Reaching the end needs at least two removals, but only one is allowed.

Constraints

• 1 ≤ m, n ≤ 40
• 1 ≤ k ≤ m · n
• Each cell is 0 or 1.
• grid[0][0] = grid[m − 1][n − 1] = 0

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“remove up to k obstacles”

→ The removals are a budget. A state has to carry how many are left.

“fewest moves”

→ Every move costs one, so BFS gives the fewest moves. The state is bigger than the cell alone.

“up to k removals”

→ More removals left is never worse. That rule lets the search skip weaker states.

Approach Complexity Ladder

Brute Force — DFS over paths, carrying the removals leftO(3^(m · n))O(m · n)

DFS over simple paths; each path carries how many obstacles it may still remove

Optimal — BFS over (cell, removals left), keep the best per cellO(m · n · k)O(m · n · k)

BFS over states (cell, removals left); skip a state that a better one already covers

Brute Force Intuition

Walk every simple path from the top-left cell, carrying the number of removals left. Stepping onto an obstacle uses one removal, and a path with none left is cut. Keep the fewest moves that reach the end. Every path is walked on its own, so the work is exponential.

Recognition rule for this sub-pattern:

Unweighted grid → BFS. Extra state (keys, obstacles removed) becomes part of the node.

Remember this · Graph BFS / DFS

Trigger

Grid of cells, islands, regions, "minimum steps", transformations between states, connected components.

Think

What is a node, what is an edge, and do I need the shortest path or just reachability?

Key question

How do I mark visited so I never process a node twice?

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