Medium
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
import kotlin.math.abs
import kotlin.math.max
class Solution {
fun isReachableAtTime(sx: Int, sy: Int, fx: Int, fy: Int, t: Int): Boolean {
if (sx == fx && sy == fy) {
return t != 1
}
val width = abs((sx - fx)) + 1
val height = abs((sy - fy)) + 1
return max(width, height) - 1 <= t
}
}