README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2800-2899/2849.Determine%20if%20a%20Cell%20Is%20Reachable%20at%20a%20Given%20Time/README_EN.md	1515	Weekly Contest 362 Q2	Math

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

中文文档

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

Solution 1: Case Discussion

If the starting point and the destination are the same, then we can only reach the destination within the given time if $t \neq 1$.

Otherwise, we can calculate the difference in the x and y coordinates between the starting point and the destination, and then take the maximum value. If the maximum value is less than or equal to the given time, then we can reach the destination within the given time.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

C#

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;
    }
}