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            "title": "Minimum Operations to Make the Array K-Increasing",
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            "content": "<p>You are given a <strong>0-indexed</strong> array <code>arr</code> consisting of <code>n</code> positive integers, and a positive integer <code>k</code>.</p>\n\n<p>The array <code>arr</code> is called <strong>K-increasing</strong> if <code>arr[i-k] &lt;= arr[i]</code> holds for every index <code>i</code>, where <code>k &lt;= i &lt;= n-1</code>.</p>\n\n<ul>\n\t<li>For example, <code>arr = [4, 1, 5, 2, 6, 2]</code> is K-increasing for <code>k = 2</code> because:\n\n\t<ul>\n\t\t<li><code>arr[0] &lt;= arr[2] (4 &lt;= 5)</code></li>\n\t\t<li><code>arr[1] &lt;= arr[3] (1 &lt;= 2)</code></li>\n\t\t<li><code>arr[2] &lt;= arr[4] (5 &lt;= 6)</code></li>\n\t\t<li><code>arr[3] &lt;= arr[5] (2 &lt;= 2)</code></li>\n\t</ul>\n\t</li>\n\t<li>However, the same <code>arr</code> is not K-increasing for <code>k = 1</code> (because <code>arr[0] &gt; arr[1]</code>) or <code>k = 3</code> (because <code>arr[0] &gt; arr[3]</code>).</li>\n</ul>\n\n<p>In one <strong>operation</strong>, you can choose an index <code>i</code> and <strong>change</strong> <code>arr[i]</code> into <strong>any</strong> positive integer.</p>\n\n<p>Return <em>the <strong>minimum number of operations</strong> required to make the array K-increasing for the given </em><code>k</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [5,4,3,2,1], k = 1\n<strong>Output:</strong> 4\n<strong>Explanation:\n</strong>For k = 1, the resultant array has to be non-decreasing.\nSome of the K-increasing arrays that can be formed are [5,<u><strong>6</strong></u>,<u><strong>7</strong></u>,<u><strong>8</strong></u>,<u><strong>9</strong></u>], [<u><strong>1</strong></u>,<u><strong>1</strong></u>,<u><strong>1</strong></u>,<u><strong>1</strong></u>,1], [<u><strong>2</strong></u>,<u><strong>2</strong></u>,3,<u><strong>4</strong></u>,<u><strong>4</strong></u>]. All of them require 4 operations.\nIt is suboptimal to change the array to, for example, [<u><strong>6</strong></u>,<u><strong>7</strong></u>,<u><strong>8</strong></u>,<u><strong>9</strong></u>,<u><strong>10</strong></u>] because it would take 5 operations.\nIt can be shown that we cannot make the array K-increasing in less than 4 operations.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [4,1,5,2,6,2], k = 2\n<strong>Output:</strong> 0\n<strong>Explanation:</strong>\nThis is the same example as the one in the problem description.\nHere, for every index i where 2 &lt;= i &lt;= 5, arr[i-2] &lt;=<b> </b>arr[i].\nSince the given array is already K-increasing, we do not need to perform any operations.</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [4,1,5,2,6,2], k = 3\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\nIndices 3 and 5 are the only ones not satisfying arr[i-3] &lt;= arr[i] for 3 &lt;= i &lt;= 5.\nOne of the ways we can make the array K-increasing is by changing arr[3] to 4 and arr[5] to 5.\nThe array will now be [4,1,5,<u><strong>4</strong></u>,6,<u><strong>5</strong></u>].\nNote that there can be other ways to make the array K-increasing, but none of them require less than 2 operations.\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;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= arr[i], k &lt;= arr.length</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int kIncreasing(vector<int>& arr, int k) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public int kIncreasing(int[] arr, int k) {\n        \n    }\n}",
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                    "code": "class Solution(object):\n    def kIncreasing(self, arr, k):\n        \"\"\"\n        :type arr: List[int]\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "class Solution:\n    def kIncreasing(self, arr: List[int], k: int) -> int:\n        ",
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                    "code": "\n\nint kIncreasing(int* arr, int arrSize, int k){\n\n}",
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                    "code": "public class Solution {\n    public int KIncreasing(int[] arr, int k) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[]} arr\n * @param {number} k\n * @return {number}\n */\nvar kIncreasing = function(arr, k) {\n    \n};",
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                    "code": "# @param {Integer[]} arr\n# @param {Integer} k\n# @return {Integer}\ndef k_increasing(arr, k)\n    \nend",
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                    "lang": "Swift",
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                    "code": "class Solution {\n    func kIncreasing(_ arr: [Int], _ k: Int) -> Int {\n        \n    }\n}",
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                    "lang": "Go",
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                    "code": "func kIncreasing(arr []int, k int) int {\n    \n}",
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                    "lang": "Scala",
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                    "code": "object Solution {\n    def kIncreasing(arr: Array[Int], k: Int): Int = {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun kIncreasing(arr: IntArray, k: Int): Int {\n        \n    }\n}",
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                    "lang": "Rust",
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                    "code": "impl Solution {\n    pub fn k_increasing(arr: Vec<i32>, k: i32) -> i32 {\n        \n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $arr\n     * @param Integer $k\n     * @return Integer\n     */\n    function kIncreasing($arr, $k) {\n        \n    }\n}",
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                    "lang": "TypeScript",
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                    "code": "function kIncreasing(arr: number[], k: number): number {\n\n};",
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                    "code": "(define/contract (k-increasing arr k)\n  (-> (listof exact-integer?) exact-integer? exact-integer?)\n\n  )",
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                    "lang": "Erlang",
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                    "code": "-spec k_increasing(Arr :: [integer()], K :: integer()) -> integer().\nk_increasing(Arr, K) ->\n  .",
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                    "code": "defmodule Solution do\n  @spec k_increasing(arr :: [integer], k :: integer) :: integer\n  def k_increasing(arr, k) do\n\n  end\nend",
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            "hints": [
                "Can we divide the array into non-overlapping subsequences and simplify the problem?",
                "In the final array, arr[i-k] ≤ arr[i] should hold. We can use this to divide the array into at most k non-overlapping sequences, where arr[i] will belong to the (i%k)th sequence.",
                "Now our problem boils down to performing the minimum operations on each sequence such that it becomes non-decreasing. Our answer will be the sum of operations on each sequence.",
                "Which indices of a sequence should we not change in order to count the minimum operations? Can finding the longest non-decreasing subsequence of the sequence help?"
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