class Solution {
public int minimumTime(int[][] grid) {
if (grid[0][1] > 1 && grid[1][0] > 1) {
return -1;
}
int m = grid.length, n = grid[0].length;
int[][] dist = new int[m][n];
for (var e : dist) {
Arrays.fill(e, 1 << 30);
}
dist[0][0] = 0;
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
pq.offer(new int[] {0, 0, 0});
int[] dirs = {-1, 0, 1, 0, -1};
while (true) {
var p = pq.poll();
int i = p[1], j = p[2];
if (i == m - 1 && j == n - 1) {
return p[0];
}
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n) {
int nt = p[0] + 1;
if (nt < grid[x][y]) {
nt = grid[x][y] + (grid[x][y] - nt) % 2;
}
if (nt < dist[x][y]) {
dist[x][y] = nt;
pq.offer(new int[] {nt, x, y});
}
}
}
}
}
}https://leetcode.com/problems/minimum-time-to-visit-a-cell-in-a-grid/?envType=daily-question&envId=2024-11-29
LeetCode
Minimum Time to Visit a Cell In a Grid - LeetCode
Can you solve this real interview question? Minimum Time to Visit a Cell In a Grid - You are given a m x n matrix grid consisting of non-negative integers where grid[row][col] represents the minimum time required to be able to visit the cell (row, col), which…
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def raqamlar_yigindisi(n):
if n < 10: #
return n
else:
return n % 10 + raqamlar_yigindisi(n // 10)
if n < 10: #
return n
else:
return n % 10 + raqamlar_yigindisi(n // 10)
Berilgan N natural sonning raqamlari yig'indisini rekursiv funksiya yordamida hisoblang.
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