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

https://leetcode.com/problems/determine-if-a-cell-is-reachable-at-a-given-time/

Author: sayeed205

Medium/2849. Determine if a Cell Is Reachable at a Given Time/2849. Determine if a Cell Is Reachable at a Given Time.java

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+class Solution {
+    public boolean isReachableAtTime(int sx, int sy, int fx, int fy, int t) {
+        int dx = Math.abs(sx - fx);
+        int dy = Math.abs(sy - fy);
+
+        if (dx == 0 && dy == 0) {
+            return t != 1;
+        } else {
+            return Math.max(dx, dy) <= t;
+        }
+    }
+}

Medium/2849. Determine if a Cell Is Reachable at a Given Time/2849. Determine if a Cell Is Reachable at a Given Time.js

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+/**
+ * Determines if a cell is reachable at a given time.
+ *
+ * @param {number} sx - Starting x-coordinate
+ * @param {number} sy - Starting y-coordinate
+ * @param {number} fx - Ending x-coordinate
+ * @param {number} fy - Ending y-coordinate
+ * @param {number} t - Time in seconds
+ * @return {boolean} - Returns true if the cell can be reached in exactly t seconds, or false otherwise
+ */
+var isReachableAtTime = function (sx, sy, fx, fy, t) {
+    const dx = Math.abs(sx - fx);
+    const dy = Math.abs(sy - fy);
+
+    if (dx === 0 && dy === 0) {
+        return t !== 1;
+    } else {
+        return Math.max(dx, dy) <= t;
+    }
+};

Medium/2849. Determine if a Cell Is Reachable at a Given Time/2849. Determine if a Cell Is Reachable at a Given Time.py

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+class Solution:
+    def isReachableAtTime(self, sx: int, sy: int, fx: int, fy: int, t: int) -> bool:
+        dx = abs(sx - fx)
+        dy = abs(sy - fy)
+
+        if dx == 0 and dy == 0:
+            return t != 1
+        else:
+            return max(dx, dy) <= t

Medium/2849. Determine if a Cell Is Reachable at a Given Time/2849. Determine if a Cell Is Reachable at a Given Time.rs

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+impl Solution {
+    pub fn is_reachable_at_time(sx: i32, sy: i32, fx: i32, fy: i32, t: i32) -> bool {
+        let dx = (sx - fx).abs();
+        let dy = (sy - fy).abs();
+        if dx == 0 && dy == 0 {
+            t != 1
+        } else {
+            dx.max(dy) <= t
+        }
+    }
+}

Medium/2849. Determine if a Cell Is Reachable at a Given Time/2849. Determine if a Cell Is Reachable at a Given Time.ts

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+function isReachableAtTime(
+    sx: number,
+    sy: number,
+    fx: number,
+    fy: number,
+    t: number
+): boolean {
+    const dx = Math.abs(sx - fx);
+    const dy = Math.abs(sy - fy);
+
+    if (dx === 0 && dy === 0) {
+        return t !== 1;
+    } else {
+        return Math.max(dx, dy) <= t;
+    }
+}

Medium/2849. Determine if a Cell Is Reachable at a Given Time/README.md

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+## [2849. Determine if a Cell Is Reachable at a Given Time](https://leetcode.com/problems/determine-if-a-cell-is-reachable-at-a-given-time/)
+
+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:
+
+![example1](https://assets.leetcode.com/uploads/2023/08/05/example2.svg)
+
+<pre>
+<strong>Input:</strong> sx = 3, sy = 1, fx = 7, fy = 3, t = 3
+<strong>Output:</strong> false
+<strong>Explanation:</strong> 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.
+</pre>
+
+#### Constraints:
+
+-   <code>1 <= sx, sy, fx, fy <= 10<sup>9</sup></code>
+-   <code>0 <= t <= 10<sup>9</sup></code>