# 2849. Determine if a Cell Is Reachable at a Given Time

## Description

You are given four integers sx, sy, fx, fy, and a non-negative integer t.

In an infinite 2D grid, you start at the cell (sx, sy). Each second, you must move to any of its adjacent cells.

Return true if you can reach cell (fx, fy) after exactly t seconds, or false otherwise.

A cell's adjacent cells are the 8 cells around it that share at least one corner with it. You can visit the same cell several times.

Example 1:

Input: sx = 2, sy = 4, fx = 7, fy = 7, t = 6
Output: true
Explanation: Starting at cell (2, 4), we can reach cell (7, 7) in exactly 6 seconds by going through the cells depicted in the picture above.


Example 2:

Input: sx = 3, sy = 1, fx = 7, fy = 3, t = 3
Output: false
Explanation: Starting at cell (3, 1), it takes at least 4 seconds to reach cell (7, 3) by going through the cells depicted in the picture above. Hence, we cannot reach cell (7, 3) at the third second.


Constraints:

• 1 <= sx, sy, fx, fy <= 109
• 0 <= t <= 109

## Solutions

• class Solution {
public boolean isReachableAtTime(int sx, int sy, int fx, int fy, int t) {
if (sx == fx && sy == fy) {
return t != 1;
}
int dx = Math.abs(sx - fx);
int dy = Math.abs(sy - fy);
return Math.max(dx, dy) <= t;
}
}

• class Solution {
public:
bool isReachableAtTime(int sx, int sy, int fx, int fy, int t) {
if (sx == fx && sy == fy) {
return t != 1;
}
int dx = abs(fx - sx), dy = abs(fy - sy);
return max(dx, dy) <= t;
}
};

• class Solution:
def isReachableAtTime(self, sx: int, sy: int, fx: int, fy: int, t: int) -> bool:
if sx == fx and sy == fy:
return t != 1
dx = abs(sx - fx)
dy = abs(sy - fy)
return max(dx, dy) <= t


• func isReachableAtTime(sx int, sy int, fx int, fy int, t int) bool {
if sx == fx && sy == fy {
return t != 1
}
dx := abs(sx - fx)
dy := abs(sy - fy)
return max(dx, dy) <= t
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• function isReachableAtTime(sx: number, sy: number, fx: number, fy: number, t: number): boolean {
if (sx === fx && sy === fy) {
return t !== 1;
}
const dx = Math.abs(sx - fx);
const dy = Math.abs(sy - fy);
return Math.max(dx, dy) <= t;
}


• public class Solution {
public bool IsReachableAtTime(int sx, int sy, int fx, int fy, int t) {
if (sx == fx && sy == fy)
return t != 1;
return Math.Max(Math.Abs(sx - fx), Math.Abs(sy - fy)) <= t;
}
}