leetcode-problemset/leetcode-cn/originData/odd-even-jump.json

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            "title": "Odd Even Jump",
            "titleSlug": "odd-even-jump",
            "content": "<p>You are given an integer array <code>arr</code>. From some starting index, you can make a series of jumps. The (1<sup>st</sup>, 3<sup>rd</sup>, 5<sup>th</sup>, ...) jumps in the series are called <strong>odd-numbered jumps</strong>, and the (2<sup>nd</sup>, 4<sup>th</sup>, 6<sup>th</sup>, ...) jumps in the series are called <strong>even-numbered jumps</strong>. Note that the <strong>jumps</strong> are numbered, not the indices.</p>\n\n<p>You may jump forward from index <code>i</code> to index <code>j</code> (with <code>i &lt; j</code>) in the following way:</p>\n\n<ul>\n\t<li>During <strong>odd-numbered jumps</strong> (i.e., jumps 1, 3, 5, ...), you jump to the index <code>j</code> such that <code>arr[i] &lt;= arr[j]</code> and <code>arr[j]</code> is the smallest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>\n\t<li>During <strong>even-numbered jumps</strong> (i.e., jumps 2, 4, 6, ...), you jump to the index <code>j</code> such that <code>arr[i] &gt;= arr[j]</code> and <code>arr[j]</code> is the largest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>\n\t<li>It may be the case that for some index <code>i</code>, there are no legal jumps.</li>\n</ul>\n\n<p>A starting index is <strong>good</strong> if, starting from that index, you can reach the end of the array (index <code>arr.length - 1</code>) by jumping some number of times (possibly 0 or more than once).</p>\n\n<p>Return <em>the number of <strong>good</strong> starting indices</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [10,13,12,14,15]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \nFrom starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more.\nFrom starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.\nFrom starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.\nFrom starting index i = 4, we have reached the end already.\nIn total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of\njumps.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [2,3,1,1,4]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> \nFrom starting index i = 0, we make jumps to i = 1, i = 2, i = 3:\nDuring our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].\nDuring our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3\nDuring our 3rd jump (odd-numbered), we jump from i = 2 to i = 3 because arr[3] is the smallest value in [arr[3], arr[4]] that is greater than or equal to arr[2].\nWe can&#39;t jump from i = 3 to i = 4, so the starting index i = 0 is not good.\nIn a similar manner, we can deduce that:\nFrom starting index i = 1, we jump to i = 4, so we reach the end.\nFrom starting index i = 2, we jump to i = 3, and then we can&#39;t jump anymore.\nFrom starting index i = 3, we jump to i = 4, so we reach the end.\nFrom starting index i = 4, we are already at the end.\nIn total, there are 3 different starting indices i = 1, i = 3, and i = 4, where we can reach the end with some\nnumber of jumps.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [5,1,3,4,2]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We can reach the end from starting indices 1, 2, and 4.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= arr.length &lt;= 2 * 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= arr[i] &lt; 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "奇偶跳",
            "translatedContent": "<p>给定一个整数数组 <code>A</code>，你可以从某一起始索引出发，跳跃一定次数。在你跳跃的过程中，第 1、3、5... 次跳跃称为奇数跳跃，而第 2、4、6... 次跳跃称为偶数跳跃。</p>\n\n<p>你可以按以下方式从索引 <code>i</code>&nbsp;向后跳转到索引 <code>j</code>（其中 <code>i &lt; j</code>）：</p>\n\n<ul>\n\t<li>在进行奇数跳跃时（如，第&nbsp;1，3，5... 次跳跃），你将会跳到索引 <code>j</code>，使得 <code>A[i] &lt;=&nbsp;A[j]</code>，<code>A[j]</code> 是可能的最小值。如果存在多个这样的索引 <code>j</code>，你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code> 上。</li>\n\t<li>在进行偶数跳跃时（如，第&nbsp;2，4，6... 次跳跃），你将会跳到索引&nbsp;<code>j</code>，使得 <code>A[i] &gt;= A[j]</code>，<code>A[j]</code> 是可能的最大值。如果存在多个这样的索引 <code>j</code>，你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code>&nbsp;上。</li>\n\t<li>（对于某些索引 <code>i</code>，可能无法进行合乎要求的跳跃。）</li>\n</ul>\n\n<p>如果从某一索引开始跳跃一定次数（可能是 0 次或多次），就可以到达数组的末尾（索引 <code>A.length - 1</code>），那么该索引就会被认为是好的起始索引。</p>\n\n<p>返回好的起始索引的数量。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>[10,13,12,14,15]\n<strong>输出：</strong>2\n<strong>解释： </strong>\n从起始索引 i = 0 出发，我们可以跳到 i = 2，（因为 A[2] 是 A[1]，A[2]，A[3]，A[4] 中大于或等于 A[0] 的最小值），然后我们就无法继续跳下去了。\n从起始索引 i = 1 和 i = 2 出发，我们可以跳到 i = 3，然后我们就无法继续跳下去了。\n从起始索引 i = 3 出发，我们可以跳到 i = 4，到达数组末尾。\n从起始索引 i = 4 出发，我们已经到达数组末尾。\n总之，我们可以从 2 个不同的起始索引（i = 3, i = 4）出发，通过一定数量的跳跃到达数组末尾。\n</pre>\n\n<p><strong>示例&nbsp;2：</strong></p>\n\n<pre><strong>输入：</strong>[2,3,1,1,4]\n<strong>输出：</strong>3\n<strong>解释：</strong>\n从起始索引 i=0 出发，我们依次可以跳到 i = 1，i = 2，i = 3：\n\n在我们的第一次跳跃（奇数）中，我们先跳到 i = 1，因为 A[1] 是（A[1]，A[2]，A[3]，A[4]）中大于或等于 A[0] 的最小值。\n\n在我们的第二次跳跃（偶数）中，我们从 i = 1 跳到 i = 2，因为 A[2] 是（A[2]，A[3]，A[4]）中小于或等于 A[1] 的最大值。A[3] 也是最大的值，但 2 是一个较小的索引，所以我们只能跳到 i = 2，而不能跳到 i = 3。\n\n在我们的第三次跳跃（奇数）中，我们从 i = 2 跳到 i = 3，因为 A[3] 是（A[3]，A[4]）中大于或等于 A[2] 的最小值。\n\n我们不能从 i = 3 跳到 i = 4，所以起始索引 i = 0 不是好的起始索引。\n\n类似地，我们可以推断：\n从起始索引 i = 1 出发， 我们跳到 i = 4，这样我们就到达数组末尾。\n从起始索引 i = 2 出发， 我们跳到 i = 3，然后我们就不能再跳了。\n从起始索引 i = 3 出发， 我们跳到 i = 4，这样我们就到达数组末尾。\n从起始索引 i = 4 出发，我们已经到达数组末尾。\n总之，我们可以从 3 个不同的起始索引（i = 1, i = 3, i = 4）出发，通过一定数量的跳跃到达数组末尾。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>[5,1,3,4,2]\n<strong>输出：</strong>3\n<strong>解释： </strong>\n我们可以从起始索引 1，2，4 出发到达数组末尾。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ol>\n\t<li><code>1 &lt;= A.length &lt;= 20000</code></li>\n\t<li><code>0 &lt;= A[i] &lt; 100000</code></li>\n</ol>\n",
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href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\\" target=\\\"_blank\\\"> datastructures-js\\/queue<\\/a>\\u3002<\\/p>\"],\"ruby\":[\"Ruby\",\"<p>\\u4f7f\\u7528<code>Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n<p>\\u4e00\\u4e9b\\u5e38\\u7528\\u7684\\u6570\\u636e\\u7ed3\\u6784\\u5df2\\u5728 Algorithms \\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\"<p>\\u7248\\u672c\\uff1a<code>Swift 5.5.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 <a href=\\\"https:\\/\\/swift.org\\/download\\/\\\" target=\\\"_blank\\\">Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\"<p>\\u7248\\u672c\\uff1a<code>Go 1.17<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 <a href=\\\"https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 4.5.4<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.3<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 24.2\"],\"elixir\":[\"Elixir\",\"Elixir 1.13.0 with Erlang\\/OTP 24.2\"]}",
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