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This approach leverages the concept of "Manhattan distance" coupled with the "parity check" to determine if the cell can be reached in exactly t steps.
The minimum number of steps required to reach from (sx, sy) to (fx, fy) in a grid is the maximum of the horizontal and vertical distances between these points, known as Chebyshev distance. If this minimal distance is greater than t, the function should return false instantly.
Moreover, if the difference between t and this minimum distance is even, the extra steps required can effectively be taken by oscillating on the grid or taking additional allowable steps. If not, reaching in t steps becomes impossible.
Time Complexity: O(1), as it involves simple arithmetic operations.
Space Complexity: O(1), as no extra space is used.
1public class Solution {
2 public bool IsCellReachable(int sx, int sy, int fx, int fy, int t) {
3 int min_steps = Math.Max(Math.Abs(fx - sx), Math.Abs(fy - sy));
4 return min_steps <= t && (t - min_steps) % 2 == 0;
5 }
6}This C# method uses Math class methods to calculate the distances and evaluate the reachability within given constraints.
This approach models the grid and implements a Breadth-First Search (BFS) to simulate traversing from the starting point to check if reaching precisely at t seconds is possible.
Instead of directly computing the distance, this method carries out a brute-force simulation of the movement, exploring all potential paths using a FIFO queue structure inherent to BFS algorithms. The goal is to maintain a count of steps and analyze pathways that may uniquely cover the grid in t steps.
Time Complexity: Highly complex, typically O(n^2).
Space Complexity: Also complex, with decisions based on array allocations with a theoretical upper limit.
1import java.util.*;
2
3public boolean isCellReachable(int sx, int sy, int fx, int fy, int t) {
4 Queue<int[]> queue = new LinkedList<>();
5 queue.offer(new int[]{sx, sy, 0});
6 int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
7
8 while (!queue.isEmpty()) {
9 int[] node = queue.poll();
10 int x = node[0], y = node[1], steps = node[2];
11
12 if (x == fx && y == fy && steps == t)
13 return true;
14 if (steps >= t)
15 continue;
16
17 for (int[] dir : dirs) {
18 queue.offer(new int[]{x + dir[0], y + dir[1], steps + 1});
19 }
20 }
21 return false;
22}Java offers efficient queue handling. The BFS runs until the end condition of the time limit is met. The number of directions broadens by 45 degrees, ensuring eight potential transitions per iteration.