{ "data": { "question": { "questionId": "3922", "questionFrontendId": "3609", "categoryTitle": "Algorithms", "boundTopicId": 3714247, "title": "Minimum Moves to Reach Target in Grid", "titleSlug": "minimum-moves-to-reach-target-in-grid", "content": "
You are given four integers sx, sy, tx, and ty, representing two points (sx, sy) and (tx, ty) on an infinitely large 2D grid.
You start at (sx, sy).
At any point (x, y), define m = max(x, y). You can either:
(x + m, y), or(x, y + m).Return the minimum number of moves required to reach (tx, ty). If it is impossible to reach the target, return -1.
\n
Example 1:
\n\nInput: sx = 1, sy = 2, tx = 5, ty = 4
\n\nOutput: 2
\n\nExplanation:
\n\nThe optimal path is:
\n\nmax(1, 2) = 2. Increase the y-coordinate by 2, moving from (1, 2) to (1, 2 + 2) = (1, 4).max(1, 4) = 4. Increase the x-coordinate by 4, moving from (1, 4) to (1 + 4, 4) = (5, 4).Thus, the minimum number of moves to reach (5, 4) is 2.
Example 2:
\n\nInput: sx = 0, sy = 1, tx = 2, ty = 3
\n\nOutput: 3
\n\nExplanation:
\n\nThe optimal path is:
\n\nmax(0, 1) = 1. Increase the x-coordinate by 1, moving from (0, 1) to (0 + 1, 1) = (1, 1).max(1, 1) = 1. Increase the x-coordinate by 1, moving from (1, 1) to (1 + 1, 1) = (2, 1).max(2, 1) = 2. Increase the y-coordinate by 2, moving from (2, 1) to (2, 1 + 2) = (2, 3).Thus, the minimum number of moves to reach (2, 3) is 3.
Example 3:
\n\nInput: sx = 1, sy = 1, tx = 2, ty = 2
\n\nOutput: -1
\n\nExplanation:
\n\n(2, 2) from (1, 1) using the allowed moves. Thus, the answer is -1.\n
Constraints:
\n\n0 <= sx <= tx <= 1090 <= sy <= ty <= 109给你四个整数 sx、sy、tx 和 ty,表示在一个无限大的二维网格上的两个点 (sx, sy) 和 (tx, ty)。
你的起点是 (sx, sy)。
在任何位置 (x, y),定义 m = max(x, y)。你可以执行以下两种操作之一:
(x + m, y),或者(x, y + m)。返回到达 (tx, ty) 所需的 最小 移动次数。如果无法到达目标点,则返回 -1。
\n\n
示例 1:
\n\n输入: sx = 1, sy = 2, tx = 5, ty = 4
\n\n输出: 2
\n\n解释:
\n\n最优路径如下:
\n\nmax(1, 2) = 2。增加 y 坐标 2,从 (1, 2) 移动到 (1, 2 + 2) = (1, 4)。max(1, 4) = 4。增加 x 坐标 4,从 (1, 4) 移动到 (1 + 4, 4) = (5, 4)。因此,到达 (5, 4) 的最小移动次数是 2。
示例 2:
\n\n输入: sx = 0, sy = 1, tx = 2, ty = 3
\n\n输出: 3
\n\n解释:
\n\n最优路径如下:
\n\nmax(0, 1) = 1。增加 x 坐标 1,从 (0, 1) 移动到 (0 + 1, 1) = (1, 1)。max(1, 1) = 1。增加 x 坐标 1,从 (1, 1) 移动到 (1 + 1, 1) = (2, 1)。max(2, 1) = 2。增加 y 坐标 2,从 (2, 1) 移动到 (2, 1 + 2) = (2, 3)。因此,到达 (2, 3) 的最小移动次数是 3。
示例 3:
\n\n输入: sx = 1, sy = 1, tx = 2, ty = 2
\n\n输出: -1
\n\n解释:
\n\n(1, 1) 到达 (2, 2)。因此,答案是 -1。\n\n
提示:
\n\n0 <= sx <= tx <= 1090 <= sy <= ty <= 109(tx, ty) to (sx, sy), undoing one move at each step.",
"If the larger coordinate >= 2 × (the smaller), undo by halving the larger; otherwise undo by subtracting the smaller from the larger.",
"Count these undo-steps until you hit (sx, sy) (return the count), or return -1 if you drop below or get stuck."
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