leetcode-problemset/leetcode-cn/originData/find-in-mountain-array.json

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        "question": {
            "questionId": "1185",
            "questionFrontendId": "1095",
            "categoryTitle": "Algorithms",
            "boundTopicId": 10005,
            "title": "Find in Mountain Array",
            "titleSlug": "find-in-mountain-array",
            "content": "<p><em>(This problem is an <strong>interactive problem</strong>.)</em></p>\n\n<p>You may recall that an array <code>arr</code> is a <strong>mountain array</strong> if and only if:</p>\n\n<ul>\n\t<li><code>arr.length &gt;= 3</code></li>\n\t<li>There exists some <code>i</code> with <code>0 &lt; i &lt; arr.length - 1</code> such that:\n\t<ul>\n\t\t<li><code>arr[0] &lt; arr[1] &lt; ... &lt; arr[i - 1] &lt; arr[i]</code></li>\n\t\t<li><code>arr[i] &gt; arr[i + 1] &gt; ... &gt; arr[arr.length - 1]</code></li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>Given a mountain array <code>mountainArr</code>, return the <strong>minimum</strong> <code>index</code> such that <code>mountainArr.get(index) == target</code>. If such an <code>index</code> does not exist, return <code>-1</code>.</p>\n\n<p><strong>You cannot access the mountain array directly.</strong> You may only access the array using a <code>MountainArray</code> interface:</p>\n\n<ul>\n\t<li><code>MountainArray.get(k)</code> returns the element of the array at index <code>k</code> (0-indexed).</li>\n\t<li><code>MountainArray.length()</code> returns the length of the array.</li>\n</ul>\n\n<p>Submissions making more than <code>100</code> calls to <code>MountainArray.get</code> will be judged <em>Wrong Answer</em>. Also, any solutions that attempt to circumvent the judge will result in disqualification.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> array = [1,2,3,4,5,3,1], target = 3\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> array = [0,1,2,4,2,1], target = 3\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> 3 does not exist in <code>the array,</code> so we return -1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= mountain_arr.length() &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= target &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= mountain_arr.get(index) &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "山脉数组中查找目标值",
            "translatedContent": "<p>（这是一个 <strong>交互式问题&nbsp;</strong>）</p>\n\n<p>给你一个 <strong>山脉数组</strong>&nbsp;<code>mountainArr</code>，请你返回能够使得&nbsp;<code>mountainArr.get(index)</code>&nbsp;<strong>等于</strong>&nbsp;<code>target</code>&nbsp;<strong>最小</strong>&nbsp;的下标 <code>index</code>&nbsp;值。</p>\n\n<p>如果不存在这样的下标 <code>index</code>，就请返回&nbsp;<code>-1</code>。</p>\n\n<p>&nbsp;</p>\n\n<p>何为山脉数组？如果数组&nbsp;<code>A</code> 是一个山脉数组的话，那它满足如下条件：</p>\n\n<p><strong>首先</strong>，<code>A.length &gt;= 3</code></p>\n\n<p><strong>其次</strong>，在&nbsp;<code>0 &lt; i&nbsp;&lt; A.length - 1</code>&nbsp;条件下，存在 <code>i</code> 使得：</p>\n\n<ul>\n\t<li><code>A[0] &lt; A[1] &lt; ... A[i-1] &lt; A[i]</code></li>\n\t<li><code>A[i] &gt; A[i+1] &gt; ... &gt; A[A.length - 1]</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p>你将&nbsp;<strong>不能直接访问该山脉数组</strong>，必须通过&nbsp;<code>MountainArray</code>&nbsp;接口来获取数据：</p>\n\n<ul>\n\t<li><code>MountainArray.get(k)</code>&nbsp;- 会返回数组中索引为<code>k</code>&nbsp;的元素（下标从 0 开始）</li>\n\t<li><code>MountainArray.length()</code>&nbsp;- 会返回该数组的长度</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>注意：</strong></p>\n\n<p>对&nbsp;<code>MountainArray.get</code>&nbsp;发起超过 <code>100</code> 次调用的提交将被视为错误答案。此外，任何试图规避判题系统的解决方案都将会导致比赛资格被取消。</p>\n\n<p>为了帮助大家更好地理解交互式问题，我们准备了一个样例 &ldquo;<strong>答案</strong>&rdquo;：<a href=\"https://leetcode-cn.com/playground/RKhe3ave\" target=\"_blank\">https://leetcode-cn.com/playground/RKhe3ave</a>，请注意这 <strong>不是一个正确答案</strong>。</p>\n\n<ol>\n</ol>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>array = [1,2,3,4,5,3,1], target = 3\n<strong>输出：</strong>2\n<strong>解释：</strong>3 在数组中出现了两次，下标分别为 2 和 5，我们返回最小的下标 2。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>array = [0,1,2,4,2,1], target = 3\n<strong>输出：</strong>-1\n<strong>解释：</strong>3 在数组中没有出现，返回 -1。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= mountain_arr.length() &lt;= 10000</code></li>\n\t<li><code>0 &lt;= target &lt;= 10^9</code></li>\n\t<li><code>0 &lt;= mountain_arr.get(index) &lt;=&nbsp;10^9</code></li>\n</ul>\n",
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                    "code": "/**\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * class MountainArray {\n *   public:\n *     int get(int index);\n *     int length();\n * };\n */\n\nclass Solution {\npublic:\n    int findInMountainArray(int target, MountainArray &mountainArr) {\n        \n    }\n};",
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                    "code": "/**\n * // This is MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * interface MountainArray {\n *     public int get(int index) {}\n *     public int length() {}\n * }\n */\n \nclass Solution {\n    public int findInMountainArray(int target, MountainArray mountainArr) {\n        \n    }\n}",
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                    "code": "# \"\"\"\n# This is MountainArray's API interface.\n# You should not implement it, or speculate about its implementation\n# \"\"\"\n#class MountainArray(object):\n#    def get(self, index):\n#        \"\"\"\n#        :type index: int\n#        :rtype int\n#        \"\"\"\n#\n#    def length(self):\n#        \"\"\"\n#        :rtype int\n#        \"\"\"\n\nclass Solution(object):\n    def findInMountainArray(self, target, mountain_arr):\n        \"\"\"\n        :type target: integer\n        :type mountain_arr: MountainArray\n        :rtype: integer\n        \"\"\"\n        ",
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                    "code": "/**\n * *********************************************************************\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * *********************************************************************\n *\n * int get(MountainArray *, int index);\n * int length(MountainArray *);\n */\n\nint findInMountainArray(int target, MountainArray* mountainArr) {\n\t\n}",
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                    "code": "/**\n * // This is MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * class MountainArray {\n *     public int Get(int index) {}\n *     public int Length() {}\n * }\n */\n\nclass Solution {\n    public int FindInMountainArray(int target, MountainArray mountainArr) {\n        \n    }\n}",
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                    "code": "/**\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * function MountainArray() {\n *     @param {number} index\n *     @return {number}\n *     this.get = function(index) {\n *         ...\n *     };\n *\n *     @return {number}\n *     this.length = function() {\n *         ...\n *     };\n * };\n */\n\n/**\n * @param {number} target\n * @param {MountainArray} mountainArr\n * @return {number}\n */\nvar findInMountainArray = function(target, mountainArr) {\n    \n};",
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                    "code": "# This is MountainArray's API interface.\n# You should not implement it, or speculate about its implementation\n# class MountainArray\n#    def get(index):\n#       \n#    end\n#\n#    def length()\n#\t\t\n#\t end\n# end\n\n# @param {int} int\n# @param {MountainArray} mountain_arr\n# @return {int}\ndef findInMountainArray(target, mountain_arr)\n    \nend",
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                    "code": "/**\n * // This is MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * interface MountainArray {\n *     public func get(_ index: Int) -> Int {}\n *     public func length() -> Int {}\n * }\n */\n\nclass Solution {\n    func findInMountainArray(_ target: Int, _ mountainArr: MountainArray) -> Int {\n        \n    }\n}",
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                    "code": "/**\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * type MountainArray struct {\n * }\n *\n * func (this *MountainArray) get(index int) int {}\n * func (this *MountainArray) length() int {}\n */\n\nfunc findInMountainArray(target int, mountainArr *MountainArray) int {\n    \n}",
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                    "code": "/**\n * // This is MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * class MountainArray {\n *     def get(index: Int): Int = {}\n *     def length(): Int = {}\n * }\n */\n\nobject Solution {\n    def findInMountainArray(value: Int, mountainArr: MountainArray): Int = {\n        \n\t}\n}",
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                    "code": "/**\n * // This is MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * class MountainArray {\n *     fun get(index: Int): Int {}\n *     fun length(): Int {}\n * }\n */\n\nclass Solution {\n    fun findInMountainArray(target: Int, mountainArr: MountainArray): Int {\n        \n    }\n}",
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                    "code": "/**\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n *  struct MountainArray;\n *  impl MountainArray {\n *     fn get(index:i32)->i32;\n *     fn length()->i32;\n * };\n */\n\nimpl Solution {\n    pub fn find_in_mountain_array(target: i32, mountainArr: &MountainArray) -> i32 {\n        \n    }\n}",
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                    "code": "/**\n * // This is the MountainArray's API interface.\n * // You should not implement it, or speculate about its implementation\n * class Master {\n *      get(index: number): number {}\n *\n *      length(): number {}\n * }\n */\n\nfunction findInMountainArray(target: number, mountainArr: MountainArray) {\n\t\n};",
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                "Based on whether A[i-1] < A[i] < A[i+1], A[i-1] < A[i] > A[i+1], or A[i-1] > A[i] > A[i+1], we are either at the left side, peak, or right side of the mountain.  We can binary search to find the peak.\r\nAfter finding the peak, we can binary search two more times to find whether the value occurs on either side of the peak."
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