What is the time complexity of Shortest Path in a Grid with Obstacles Elimination?

The optimal solution has a time complexity of O(m*n*k).

What data structures are used to solve Shortest Path in a Grid with Obstacles Elimination?

This problem uses Array, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Shortest Path in a Grid with Obstacles Elimination?

Insight 1: BFS is suitable for finding the shortest path in an unweighted graph (or a grid where each move has a cost of 1). Insight 2: We need to track the remaining obstacle elimination capacity ('k_left') at each cell we visit. This is because reaching the same cell with different 'k_left' values represents different states.

Which companies ask LeetCode 1293. Shortest Path in a Grid with Obstacles Elimination in interviews?

This problem is frequently asked at AppFolio, Databricks, IMC, Nuro, Pinterest and 1 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Shortest Path in a Grid with Obstacles Elimination - LeetCode 1293 Solution

Q: What is the space complexity of Shortest Path in a Grid with Obstacles Elimination?

The space complexity is O(m*n*k).

Shortest Path in a Grid with Obstacles Elimination - Complete Solution Guide

Shortest Path in a Grid with Obstacles Elimination is LeetCode problem 1293, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 shor

Detailed Explanation

The problem asks us to find the shortest path from the top-left corner (0, 0) to the bottom-right corner (m-1, n-1) of a grid. The grid contains empty cells (0) and obstacles (1). We are allowed to eliminate at most 'k' obstacles. The path can only move up, down, left, or right. If a path exists within the obstacle elimination limit, we return the minimum number of steps; otherwise, we return -1.

Solution Approach

The provided code uses Breadth-First Search (BFS) to explore the grid. Each state in the BFS queue contains the number of steps taken so far, the current row and column coordinates, and the number of obstacles we can still eliminate. A 'visited' array is used to keep track of the maximum 'k_left' value with which we've visited a cell. This helps to avoid revisiting cells unnecessarily and prevents infinite loops.

Step-by-Step Algorithm

Step 1: Initialize a queue with the starting state (0 steps, row 0, column 0, k remaining eliminations).
Step 2: Initialize a 'visited' array to store the maximum remaining 'k' value with which a cell has been visited. Set visited[0][0] to k.
Step 3: While the queue is not empty:
Step 4: Dequeue a state from the queue (steps, r, c, k_left).
Step 5: If the current cell is the destination (bottom-right corner), return the 'steps'.
Step 6: For each possible move (up, down, left, right):
Step 7: Calculate the new row and column coordinates (nr, nc).
Step 8: If the new coordinates are within the grid boundaries:
Step 9: Calculate the remaining obstacle eliminations after moving to the new cell (new_k = k_left - grid[nr][nc]). If the new cell is an obstacle (grid[nr][nc] == 1), new_k will decrease by 1.
Step 10: If new_k is non-negative (we have enough eliminations) and new_k is greater than the current value in visited[nr][nc]:
Step 11: Update visited[nr][nc] with new_k.
Step 12: Enqueue the new state (steps + 1, nr, nc, new_k).
Step 13: If the queue becomes empty without finding the destination, return -1 (no path exists).

Key Insights

Insight 1: BFS is suitable for finding the shortest path in an unweighted graph (or a grid where each move has a cost of 1).
Insight 2: We need to track the remaining obstacle elimination capacity ('k_left') at each cell we visit. This is because reaching the same cell with different 'k_left' values represents different states.
Insight 3: Using a visited array storing the maximum remaining k value is crucial for optimization. We only revisit a cell if we can reach it with a higher remaining 'k' value than previously encountered.

Complexity Analysis

Time Complexity: O(m*n*k)

Space Complexity: O(m*n*k)

Topics

This problem involves: Array, Breadth-First Search, Matrix.

Companies

Asked at: AppFolio, Databricks, IMC, Nuro, Pinterest, Snap.

Shortest Path in a Grid with Obstacles Elimination | LeetCode 1293

Hard

Array Breadth-First Search Matrix

Time: O(m*n*k)Space: O(m*n*k)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;

class Solution {
    public int shortestPath(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;

        if (k >= m + n - 3) {
            return m + n - 2;
        }

        Deque<int[]> q = new ArrayDeque<>(); // (steps, r, c, k_left)
        q.offer(new int[] {0, 0, 0, k});
        int[][] visited = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                visited[i][j] = -1;
            }
        }
        visited[0][0] = k;

        while (!q.isEmpty()) {
            int[] curr = q.poll();
            int steps = curr[0];
            int r = curr[1];
            int c = curr[2];
            int k_left = curr[3];

            if (r == m - 1 && c == n - 1) {
                return steps;
            }

            int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (int[] dir : directions) {
                int dr = dir[0];
                int dc = dir[1];
                int nr = r + dr;
                int nc = c + dc;

                if (0 <= nr && nr < m && 0 <= nc && nc < n) {
                    int new_k = k_left - grid[nr][nc];

                    if (new_k >= 0 && new_k > visited[nr][nc]) {
                        visited[nr][nc] = new_k;
                        q.offer(new int[] {steps + 1, nr, nc, new_k});
                    }
                }
            }
        }

        return -1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1293: Shortest Path in a Grid with Obstacles Elimination - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Matrix

Companies

AppFolioDatabricksIMCNuroPinterestSnap

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:1293

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Initialize a queue with the starting state (0 steps, row 0, column 0, k remaining eliminations).
2
Step 2: Initialize a 'visited' array to store the maximum remaining 'k' value with which a cell has been visited. Set visited[0][0] to k.
3
Step 3: While the queue is not empty:
4
Step 4: Dequeue a state from the queue (steps, r, c, k_left).
5
Step 5: If the current cell is the destination (bottom-right corner), return the 'steps'.
6
Step 6: For each possible move (up, down, left, right):
7
Step 7: Calculate the new row and column coordinates (nr, nc).
8
Step 8: If the new coordinates are within the grid boundaries:
9
Step 9: Calculate the remaining obstacle eliminations after moving to the new cell (new_k = k_left - grid[nr][nc]). If the new cell is an obstacle (grid[nr][nc] == 1), new_k will decrease by 1.
10
Step 10: If new_k is non-negative (we have enough eliminations) and new_k is greater than the current value in visited[nr][nc]:
11
Step 11: Update visited[nr][nc] with new_k.
12
Step 12: Enqueue the new state (steps + 1, nr, nc, new_k).
13
Step 13: If the queue becomes empty without finding the destination, return -1 (no path exists).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*k)

💾

Space Complexity

O(m*n*k)

📝

Why?

The time complexity is O(m*n*k) because in the worst case, we might visit each cell (m*n) with different remaining k values (up to k). The space complexity is O(m*n*k) because the queue and the visited array can store up to m*n*k states.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A* Search

A* search is another pathfinding algorithm that can be used for this problem. It uses a heuristic function to estimate the distance to the goal, which can potentially lead to faster exploration than BFS in some cases.

Time: O(E log V) on average, where E is the number of edges and V is the number of vertices (cells)Space: O(V)

Trade-offs: A* search can be faster than BFS with a good heuristic, but the heuristic must be admissible (never overestimate the distance to the goal) to guarantee finding the optimal path. The implementation is also more complex.

✨

Key Insights

1Insight 1: BFS is suitable for finding the shortest path in an unweighted graph (or a grid where each move has a cost of 1).
2Insight 2: We need to track the remaining obstacle elimination capacity ('k_left') at each cell we visit. This is because reaching the same cell with different 'k_left' values represents different states.
3Insight 3: Using a visited array storing the maximum remaining k value is crucial for optimization. We only revisit a cell if we can reach it with a higher remaining 'k' value than previously encountered.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: k is large enough to eliminate all obstacles in a direct path: The solution includes an optimization: `if k >= m + n - 3: return m + n - 2`. This efficiently handles cases where k is large enough to eliminate all obstacles along the shortest possible path, returning the Manhattan distance directly.
Edge case 2: No path exists: The BFS algorithm will explore all possible paths within the obstacle elimination limit. If the queue becomes empty without finding the destination, the function returns -1, indicating that no path exists.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not tracking the remaining obstacle elimination capacity: A simple BFS without considering 'k_left' would likely fail because it would not be able to distinguish between different paths to the same cell with different elimination capacities.

✗Common mistake 2: Not using a 'visited' array (or using it incorrectly): Without a proper 'visited' array, the BFS could enter an infinite loop by revisiting cells unnecessarily. The 'visited' array must store the maximum remaining 'k' with which a cell has been visited to efficiently prune the search space.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning for robots in environments with obstacles: This algorithm can be applied to navigate robots through environments with obstacles, where the robot has a limited capacity to overcome obstacles (e.g., pushing objects, climbing over small barriers).

🏦Real-world scenario 2: Network routing with limited fault tolerance: In network routing, this algorithm can be used to find the shortest path between two nodes, where the 'obstacles' represent network failures and 'k' represents the number of allowed failures along the path.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?