leetcode-problemset/leetcode-cn/originData/find-nearest-point-that-has-the-same-x-or-y-coordinate.json

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        "question": {
            "questionId": "1888",
            "questionFrontendId": "1779",
            "categoryTitle": "Algorithms",
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            "title": "Find Nearest Point That Has the Same X or Y Coordinate",
            "titleSlug": "find-nearest-point-that-has-the-same-x-or-y-coordinate",
            "content": "<p>You are given two integers, <code>x</code> and <code>y</code>, which represent your current location on a Cartesian grid: <code>(x, y)</code>. You are also given an array <code>points</code> where each <code>points[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents that a point exists at <code>(a<sub>i</sub>, b<sub>i</sub>)</code>. A point is <strong>valid</strong> if it shares the same x-coordinate or the same y-coordinate as your location.</p>\n\n<p>Return <em>the index <strong>(0-indexed)</strong> of the <strong>valid</strong> point with the smallest <strong>Manhattan distance</strong> from your current location</em>. If there are multiple, return <em>the valid point with the <strong>smallest</strong> index</em>. If there are no valid points, return <code>-1</code>.</p>\n\n<p>The <strong>Manhattan distance</strong> between two points <code>(x<sub>1</sub>, y<sub>1</sub>)</code> and <code>(x<sub>2</sub>, y<sub>2</sub>)</code> is <code>abs(x<sub>1</sub> - x<sub>2</sub>) + abs(y<sub>1</sub> - y<sub>2</sub>)</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 3, y = 4, points = [[3,4]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The answer is allowed to be on the same location as your current location.</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 3, y = 4, points = [[2,3]]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> There are no valid points.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= points.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>points[i].length == 2</code></li>\n\t<li><code>1 &lt;= x, y, a<sub>i</sub>, b<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "找到最近的有相同 X 或 Y 坐标的点",
            "translatedContent": "<p>给你两个整数&nbsp;<code>x</code> 和&nbsp;<code>y</code>&nbsp;，表示你在一个笛卡尔坐标系下的&nbsp;<code>(x, y)</code>&nbsp;处。同时，在同一个坐标系下给你一个数组&nbsp;<code>points</code>&nbsp;，其中&nbsp;<code>points[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>&nbsp;表示在&nbsp;<code>(a<sub>i</sub>, b<sub>i</sub>)</code>&nbsp;处有一个点。当一个点与你所在的位置有相同的 <code>x</code> 坐标或者相同的 <code>y</code> 坐标时，我们称这个点是 <b>有效的</b>&nbsp;。</p>\n\n<p>请返回距离你当前位置&nbsp;<strong>曼哈顿距离</strong>&nbsp;最近的&nbsp;<strong>有效</strong>&nbsp;点的下标（下标从 <strong>0</strong> 开始）。如果有多个最近的有效点，请返回下标&nbsp;<strong>最小</strong>&nbsp;的一个。如果没有有效点，请返回&nbsp;<code>-1</code>&nbsp;。</p>\n\n<p>两个点 <code>(x<sub>1</sub>, y<sub>1</sub>)</code>&nbsp;和 <code>(x<sub>2</sub>, y<sub>2</sub>)</code>&nbsp;之间的 <strong>曼哈顿距离</strong>&nbsp;为&nbsp;<code>abs(x<sub>1</sub> - x<sub>2</sub>) + abs(y<sub>1</sub> - y<sub>2</sub>)</code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]\n<b>输出：</b>2\n<b>解释：</b>所有点中，[3,1]，[2,4] 和 [4,4] 是有效点。有效点中，[2,4] 和 [4,4] 距离你当前位置的曼哈顿距离最小，都为 1 。[2,4] 的下标最小，所以返回 2 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>x = 3, y = 4, points = [[3,4]]\n<b>输出：</b>0\n<b>提示：</b>答案可以与你当前所在位置坐标相同。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>x = 3, y = 4, points = [[2,3]]\n<b>输出：</b>-1\n<b>解释：</b>没有 有效点。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= points.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>points[i].length == 2</code></li>\n\t<li><code>1 &lt;= x, y, a<sub>i</sub>, b<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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