Problem
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
Solution
/**
* @param {number} sx
* @param {number} sy
* @param {number} fx
* @param {number} fy
* @param {number} t
* @return {boolean}
*/
var isReachableAtTime = function(sx, sy, fx, fy, t) {
const x = Math.abs(sx - fx);
const y = Math.abs(sy - fy);
if (sx === fx && sy === fy && t === 1) {
return false;
}
return Math.max(x, y) <= t;
};
Explain:
nope.
Complexity:
- Time complexity : O(1).
- Space complexity : O(1).