LeetCode 2577: Minimum Time to Visit a Cell In a Grid Solution
Master LeetCode problem 2577 (Minimum Time to Visit a Cell In a Grid), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2577. Minimum Time to Visit a Cell In a Grid
Problem Explanation
Explanation:
To solve this problem, we can use Dijkstra's algorithm with a priority queue. We start from the top-left cell and iterate through all possible moves to find the minimum time required to reach the bottom-right cell. We maintain a priority queue to explore cells based on the minimum time taken to reach them. We update the time in each cell if it is less than the current time plus the time taken to move to that cell. :
Solution Code
import java.util.*;
class Solution {
public int minTimeToVisitAllCells(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[2] - b[2]);
pq.offer(new int[]{0, 0, 0});
boolean[][] visited = new boolean[m][n];
visited[0][0] = true;
while (!pq.isEmpty()) {
int[] curr = pq.poll();
int row = curr[0], col = curr[1], time = curr[2];
if (row == m - 1 && col == n - 1) {
return time;
}
for (int[] dir : directions) {
int newRow = row + dir[0];
int newCol = col + dir[1];
if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n && !visited[newRow][newCol]) {
visited[newRow][newCol] = true;
int newTime = Math.max(time, grid[newRow][newCol]);
pq.offer(new int[]{newRow, newCol, newTime});
}
}
}
return -1;
}
}Try It Yourself
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