leetcode-problemset/leetcode/originData/minimum-reverse-operations.json

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        "question": {
            "questionId": "2726",
            "questionFrontendId": "2612",
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            "title": "Minimum Reverse Operations",
            "titleSlug": "minimum-reverse-operations",
            "content": "<p>You are given an integer <code>n</code> and an integer <code>p</code> in the range <code>[<font face=\"monospace\">0</font>, n - 1]</code>. Representing a <strong>0-indexed</strong> array <code>arr</code>&nbsp;of length <code>n</code> where all positions are set to <code>0</code>&#39;s, except position <code>p</code> which is set to <code>1</code>.</p>\n\n<p>You are also given an integer array <code>banned</code> containing some positions from the array. For the <strong>i</strong><sup><strong>th</strong></sup> position in <code>banned</code>, <code>arr[banned[i]] = 0</code>, and <code>banned[i] != p</code>.</p>\n\n<p>You can perform <strong>multiple</strong> operations on <code>arr</code>. In an operation, you can choose a <strong>subarray</strong> with size <code>k</code> and <strong>reverse</strong> the subarray. However, the <code>1</code> in <code>arr</code> should never go to any of the positions in <code>banned</code>. In other words, after each operation <code>arr[banned[i]]</code> <strong>remains</strong> <code>0</code>.</p>\n\n<p><em>Return an array</em> <code>ans</code> <em>where</em><em> for each </em><code>i</code><em> from </em><code>[0, n - 1]</code>, <code>ans[i]</code> <em>is the <strong>minimum</strong> number of reverse operations needed to bring the</em> <code>1</code> <em>to position</em> <code>i</code><em> in arr</em>, <em>or</em> <code>-1</code> <em>if it is impossible</em>.</p>\n\n<ul>\n\t<li>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</li>\n\t<li>The values of <code>ans[i]</code> are independent for all <code>i</code>&#39;s.</li>\n\t<li>The <strong>reverse </strong>of an array is an array containing the values in <strong>reverse order</strong>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4, p = 0, banned = [1,2], k = 4\n<strong>Output:</strong> [0,-1,-1,1]\n<strong>Explanation:</strong> In this case <code>k = 4</code> so there is only one possible reverse operation we can perform, which is reversing the whole array. Initially, 1<strong> </strong>is placed at position 0 so the amount of operations we need for position 0 is <code>0</code>. We can never place a 1 on the banned positions, so the answer for positions 1 and 2 is <code>-1</code>. Finally, with one reverse operation we can bring the 1 to index 3, so the answer for position 3 is <code>1</code>. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5, p = 0, banned = [2,4], k = 3\n<strong>Output:</strong> [0,-1,-1,-1,-1]\n<strong>Explanation:</strong> In this case the 1 is initially at position 0, so the answer for that position is <code>0</code>. We can perform reverse operations of size 3. The 1 is currently located at position 0, so we need to reverse the subarray <code>[0, 2]</code> for it to leave that position, but reversing that subarray makes position 2 have a 1, which shouldn&#39;t happen. So, we can&#39;t move the 1 from position 0, making the result for all the other positions <code>-1</code>. \n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4, p = 2, banned = [0,1,3], k = 1\n<strong>Output:</strong> [-1,-1,0,-1]\n<strong>Explanation:</strong> In this case we can only perform reverse operations of size 1.<strong>&nbsp;</strong>So the 1 never changes its position.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= p &lt;= n - 1</code></li>\n\t<li><code>0 &lt;= banned.length &lt;= n - 1</code></li>\n\t<li><code>0 &lt;= banned[i] &lt;= n - 1</code></li>\n\t<li><code>1 &lt;= k &lt;= n&nbsp;</code></li>\n\t<li><code>banned[i] != p</code></li>\n\t<li>all values in <code>banned</code>&nbsp;are <strong>unique</strong>&nbsp;</li>\n</ul>\n",
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            "difficulty": "Hard",
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                {
                    "lang": "C++",
                    "langSlug": "cpp",
                    "code": "class Solution {\npublic:\n    vector<int> minReverseOperations(int n, int p, vector<int>& banned, int k) {\n        \n    }\n};",
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                    "lang": "Java",
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                    "code": "class Solution {\n    public int[] minReverseOperations(int n, int p, int[] banned, int k) {\n        \n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Python",
                    "langSlug": "python",
                    "code": "class Solution(object):\n    def minReverseOperations(self, n, p, banned, k):\n        \"\"\"\n        :type n: int\n        :type p: int\n        :type banned: List[int]\n        :type k: int\n        :rtype: List[int]\n        \"\"\"\n        ",
                    "__typename": "CodeSnippetNode"
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                    "lang": "Python3",
                    "langSlug": "python3",
                    "code": "class Solution:\n    def minReverseOperations(self, n: int, p: int, banned: List[int], k: int) -> List[int]:\n        ",
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                {
                    "lang": "C",
                    "langSlug": "c",
                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* minReverseOperations(int n, int p, int* banned, int bannedSize, int k, int* returnSize){\n\n}",
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                    "lang": "C#",
                    "langSlug": "csharp",
                    "code": "public class Solution {\n    public int[] MinReverseOperations(int n, int p, int[] banned, int k) {\n        \n    }\n}",
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                {
                    "lang": "JavaScript",
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                    "code": "/**\n * @param {number} n\n * @param {number} p\n * @param {number[]} banned\n * @param {number} k\n * @return {number[]}\n */\nvar minReverseOperations = function(n, p, banned, k) {\n    \n};",
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                    "lang": "Ruby",
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                    "code": "# @param {Integer} n\n# @param {Integer} p\n# @param {Integer[]} banned\n# @param {Integer} k\n# @return {Integer[]}\ndef min_reverse_operations(n, p, banned, k)\n    \nend",
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                    "lang": "Swift",
                    "langSlug": "swift",
                    "code": "class Solution {\n    func minReverseOperations(_ n: Int, _ p: Int, _ banned: [Int], _ k: Int) -> [Int] {\n        \n    }\n}",
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                {
                    "lang": "Go",
                    "langSlug": "golang",
                    "code": "func minReverseOperations(n int, p int, banned []int, k int) []int {\n    \n}",
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                    "lang": "Scala",
                    "langSlug": "scala",
                    "code": "object Solution {\n    def minReverseOperations(n: Int, p: Int, banned: Array[Int], k: Int): Array[Int] = {\n        \n    }\n}",
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                    "lang": "Kotlin",
                    "langSlug": "kotlin",
                    "code": "class Solution {\n    fun minReverseOperations(n: Int, p: Int, banned: IntArray, k: Int): IntArray {\n        \n    }\n}",
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                {
                    "lang": "Rust",
                    "langSlug": "rust",
                    "code": "impl Solution {\n    pub fn min_reverse_operations(n: i32, p: i32, banned: Vec<i32>, k: i32) -> Vec<i32> {\n        \n    }\n}",
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                    "lang": "PHP",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $n\n     * @param Integer $p\n     * @param Integer[] $banned\n     * @param Integer $k\n     * @return Integer[]\n     */\n    function minReverseOperations($n, $p, $banned, $k) {\n        \n    }\n}",
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                    "lang": "TypeScript",
                    "langSlug": "typescript",
                    "code": "function minReverseOperations(n: number, p: number, banned: number[], k: number): number[] {\n\n};",
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                    "lang": "Racket",
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                    "code": "(define/contract (min-reverse-operations n p banned k)\n  (-> exact-integer? exact-integer? (listof exact-integer?) exact-integer? (listof exact-integer?))\n\n  )",
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                    "lang": "Erlang",
                    "langSlug": "erlang",
                    "code": "-spec min_reverse_operations(N :: integer(), P :: integer(), Banned :: [integer()], K :: integer()) -> [integer()].\nmin_reverse_operations(N, P, Banned, K) ->\n  .",
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                    "lang": "Elixir",
                    "langSlug": "elixir",
                    "code": "defmodule Solution do\n  @spec min_reverse_operations(n :: integer, p :: integer, banned :: [integer], k :: integer) :: [integer]\n  def min_reverse_operations(n, p, banned, k) do\n\n  end\nend",
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                {
                    "lang": "Dart",
                    "langSlug": "dart",
                    "code": "class Solution {\n  List<int> minReverseOperations(int n, int p, List<int> banned, int k) {\n\n  }\n}",
                    "__typename": "CodeSnippetNode"
                }
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            "hints": [
                "Can we use a breadth-first search to find the minimum number of operations?",
                "Find the beginning and end indices of the subarray of size k that can be reversed to bring 1 to a particular position.",
                "Can we visit every index or do we need to consider the parity of k?"
            ],
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