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:
![](https://spcdn.pages.dev/leetcode/problems/2849.Determine%20if%20a%20Cell%20Is%20Reachable%20at%20a%20Given%20Time/images/example2.svg)
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:
![](https://spcdn.pages.dev/leetcode/problems/2849.Determine%20if%20a%20Cell%20Is%20Reachable%20at%20a%20Given%20Time/images/example1.svg)
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#
Solution 1#
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| 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
|
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| 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;
}
}
|