mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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179 lines
26 KiB
JSON
179 lines
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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] <= arr[i]</code> holds for every index <code>i</code>, where <code>k <= i <= 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] <= arr[2] (4 <= 5)</code></li>\n\t\t<li><code>arr[1] <= arr[3] (1 <= 2)</code></li>\n\t\t<li><code>arr[2] <= arr[4] (5 <= 6)</code></li>\n\t\t<li><code>arr[3] <= arr[5] (2 <= 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] > arr[1]</code>) or <code>k = 3</code> (because <code>arr[0] > 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> </p>\n<p><strong class=\"example\">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 class=\"example\">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 <= i <= 5, arr[i-2] <=<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 class=\"example\">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] <= arr[i] for 3 <= i <= 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> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= arr.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= arr[i], k <= arr.length</code></li>\n</ul>\n",
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"translatedTitle": "使数组 K 递增的最少操作次数",
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"translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始包含 <code>n</code> 个正整数的数组 <code>arr</code> ，和一个正整数 <code>k</code> 。</p>\n\n<p>如果对于每个满足 <code>k <= i <= n-1</code> 的下标 <code>i</code> ，都有 <code>arr[i-k] <= arr[i]</code> ，那么我们称 <code>arr</code> 是 <strong>K</strong> <strong>递增</strong> 的。</p>\n\n<ul>\n\t<li>比方说，<code>arr = [4, 1, 5, 2, 6, 2]</code> 对于 <code>k = 2</code> 是 K 递增的，因为：\n\n\t<ul>\n\t\t<li><code>arr[0] <= arr[2] (4 <= 5)</code></li>\n\t\t<li><code>arr[1] <= arr[3] (1 <= 2)</code></li>\n\t\t<li><code>arr[2] <= arr[4] (5 <= 6)</code></li>\n\t\t<li><code>arr[3] <= arr[5] (2 <= 2)</code></li>\n\t</ul>\n\t</li>\n\t<li>但是，相同的数组 <code>arr</code> 对于 <code>k = 1</code> 不是 K 递增的（因为 <code>arr[0] > arr[1]</code>），对于 <code>k = 3</code> 也不是 K 递增的（因为 <code>arr[0] > arr[3]</code> ）。</li>\n</ul>\n\n<p>每一次 <strong>操作</strong> 中，你可以选择一个下标 <code>i</code> 并将 <code>arr[i]</code> <strong>改成任意 </strong>正整数。</p>\n\n<p>请你返回对于给定的 <code>k</code> ，使数组变成 K 递增的 <strong>最少操作次数</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>arr = [5,4,3,2,1], k = 1\n<b>输出：</b>4\n<strong>解释：\n</strong>对于 k = 1 ，数组最终必须变成非递减的。\n可行的 K 递增结果数组为 [5,<em><strong>6</strong></em>,<em><strong>7</strong></em>,<em><strong>8</strong></em>,<em><strong>9</strong></em>]，[<em><strong>1</strong></em>,<em><strong>1</strong></em>,<em><strong>1</strong></em>,<em><strong>1</strong></em>,1]，[<em><strong>2</strong></em>,<em><strong>2</strong></em>,3,<em><strong>4</strong></em>,<em><strong>4</strong></em>] 。它们都需要 4 次操作。\n次优解是将数组变成比方说 [<em><strong>6</strong></em>,<em><strong>7</strong></em>,<em><strong>8</strong></em>,<em><strong>9</strong></em>,<em><strong>10</strong></em>] ，因为需要 5 次操作。\n显然我们无法使用少于 4 次操作将数组变成 K 递增的。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>arr = [4,1,5,2,6,2], k = 2\n<b>输出：</b>0\n<strong>解释：</strong>\n这是题目描述中的例子。\n对于每个满足 2 <= i <= 5 的下标 i ，有 arr[i-2] <=<b> </b>arr[i] 。\n由于给定数组已经是 K 递增的，我们不需要进行任何操作。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>arr = [4,1,5,2,6,2], k = 3\n<b>输出：</b>2\n<strong>解释：</strong>\n下标 3 和 5 是仅有的 3 <= i <= 5 且不满足 arr[i-3] <= arr[i] 的下标。\n将数组变成 K 递增的方法之一是将 arr[3] 变为 4 ，且将 arr[5] 变成 5 。\n数组变为 [4,1,5,<em><strong>4</strong></em>,6,<em><strong>5</strong></em>] 。\n可能有其他方法将数组变为 K 递增的，但没有任何一种方法需要的操作次数小于 2 次。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= arr.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= arr[i], k <= arr.length</code></li>\n</ul>\n",
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"Can we divide the array into non-overlapping subsequences and simplify the problem?",
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"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.",
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"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.",
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