707518. Shortest Path by Removing K walls
Here is a detailed Markdown blog post for the problem "Shortest Path by Removing K walls":
Shortest Path by Removing K walls
Summary
Given an n x m grid filled with walls (0) and empty spaces (1), find the shortest path from the top-left corner to the bottom-right corner by removing at most K walls. The problem involves finding the shortest path in a grid, which is typically solved using BFS or Dijkstra's algorithm.
Detailed Explanation
The approach to solve this problem is by using Breadth-First Search (BFS). We create a queue and enqueue all possible movements from the top-left corner (up, down, left, right). We keep track of visited cells and maintain a dist array to store the shortest distance from the top-left corner. We also use a min Walls variable to store the minimum number of walls that need to be removed.
Here's a step-by-step breakdown:
- Initialize the queue with the top-left cell.
- Dequeue all cells and for each dequeued cell, explore its neighbors (up, down, left, right).
- If a neighbor is not visited before, mark it as visited and enqueue it.
- For each dequeued cell, calculate the new distance by adding 1 to the current shortest distance.
- If the new distance is less than the
distvalue at the corresponding position in thedistarray, update thedistvalue. - Repeat steps 2-5 until the queue becomes empty.
- After BFS traversal, find the minimum number of walls that need to be removed by traversing from the bottom-right corner to the top-left corner and removing the maximum number of walls that do not break the shortest path.
Time complexity: O(nm), where n is the number of rows and m is the number of columns in the grid. This is because we are visiting each cell once.
Space complexity: O(nm), for storing the dist array and visited cells.
Optimized Solutions
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
int[][] grid = {{0, 1}, {1, 0}};
int n = grid.length;
int m = grid[0].length;
int K = 2;
boolean[][] visited = new boolean[n][m];
int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
Queue<int[]> queue = new LinkedList<>();
int[] start = {0, 0};
queue.add(start);
visited[0][0] = true;
int[][] dist = new int[n][m];
for (int i = 0; i < n; i++) {
Arrays.fill(dist[i], Integer.MAX_VALUE);
}
dist[0][0] = 0;
while (!queue.isEmpty()) {
int[] curr = queue.poll();
for (int[] d : dir) {
int x = curr[0] + d[0];
int y = curr[1] + d[1];
if (x < 0 || x >= n || y < 0 || y >= m || visited[x][y]) continue;
queue.add(new int[]{x, y});
visited[x][y] = true;
dist[x][y] = Math.min(dist[x][y], dist[curr[0]][curr[1]] + 1);
}
}
// Calculate the minimum number of walls that need to be removed
int minWalls = 0;
for (int i = n - 1; i >= 0; i--) {
for (int j = m - 1; j >= 0; j--) {
if (grid[i][j] == 1 && dist[i][j] > K) minWalls++;
}
}
System.out.println("Minimum number of walls that need to be removed: " + minWalls);
}
}