Sponsored
Sponsored
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.
1#include <iostream>
2#include <cmath>
3
4bool isCellReachable(int sx, int sy, int fx, int fy, int t) {
5 int min_steps = std::max(std::abs(fx - sx), std::abs(fy - sy));
6 return (min_steps <= t && (t - min_steps) % 2 == 0);
7}Using the C++ standard library functions, the approach checks the minimum steps and the parity between t and these steps. It then returns the boolean value based on this condition.
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.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public bool IsCellReachable(int sx, int sy, int fx, int fy, int t) {
6 Queue<(int, int, int)> queue = new Queue<(int, int, int)>();
7 queue.Enqueue((sx, sy, 0));
8 int[][] directions = new int[][]
9 {
10 new int[] {-1, 0}, new int[] {1, 0}, new int[] {0, -1}, new int[] {0, 1},
11 new int[] {-1, -1}, new int[] {-1, 1}, new int[] {1, -1}, new int[] {1, 1}
12 };
13
14 while (queue.Count > 0) {
15 var (x, y, steps) = queue.Dequeue();
16
17 if (x == fx && y == fy && steps == t)
18 return true;
19 if (steps >= t)
20 continue;
21
22 foreach (var dir in directions) {
23 queue.Enqueue((x + dir[0], y + dir[1], steps + 1));
24 }
25 }
26 return false;
27 }
28}The C# design similarly follows a queue-based BFS formulation. Each tuple represents a position on the grid and the elapsed time count.