leetcode-problemset/leetcode-cn/originData/minimum-cost-to-make-array-equalindromic.json

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            "title": "Minimum Cost to Make Array Equalindromic",
            "titleSlug": "minimum-cost-to-make-array-equalindromic",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> having length <code>n</code>.</p>\n\n<p>You are allowed to perform a special move <strong>any</strong> number of times (<strong>including zero</strong>) on <code>nums</code>. In one <strong>special</strong> <strong>move</strong> you perform the following steps <strong>in order</strong>:</p>\n\n<ul>\n\t<li>Choose an index <code>i</code> in the range <code>[0, n - 1]</code>, and a <strong>positive</strong> integer <code>x</code>.</li>\n\t<li>Add <code>|nums[i] - x|</code> to the total cost.</li>\n\t<li>Change the value of <code>nums[i]</code> to <code>x</code>.</li>\n</ul>\n\n<p>A <strong>palindromic number</strong> is a positive integer that remains the same when its digits are reversed. For example, <code>121</code>, <code>2552</code> and <code>65756</code> are palindromic numbers whereas <code>24</code>, <code>46</code>, <code>235</code> are not palindromic numbers.</p>\n\n<p>An array is considered <strong>equalindromic</strong> if all the elements in the array are equal to an integer <code>y</code>, where <code>y</code> is a <strong>palindromic number</strong> less than <code>10<sup>9</sup></code>.</p>\n\n<p>Return <em>an integer denoting the <strong>minimum</strong> possible total cost to make </em><code>nums</code><em> <strong>equalindromic</strong> by performing any number of special moves.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4,5]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 3 which is a palindromic number. The cost of changing the array to [3,3,3,3,3] using 4 special moves is given by |1 - 3| + |2 - 3| + |4 - 3| + |5 - 3| = 6.\nIt can be shown that changing all elements to any palindromic number other than 3 cannot be achieved at a lower cost.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [10,12,13,14,15]\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 11 which is a palindromic number. The cost of changing the array to [11,11,11,11,11] using 5 special moves is given by |10 - 11| + |12 - 11| + |13 - 11| + |14 - 11| + |15 - 11| = 11.\nIt can be shown that changing all elements to any palindromic number other than 11 cannot be achieved at a lower cost.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [22,33,22,33,22]\n<strong>Output:</strong> 22\n<strong>Explanation:</strong> We can make the array equalindromic by changing all elements to 22 which is a palindromic number. The cost of changing the array to [22,22,22,22,22] using 2 special moves is given by |33 - 22| + |33 - 22| = 22.\nIt can be shown that changing all elements to any palindromic number other than 22 cannot be achieved at a lower cost.\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>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "使数组成为等数数组的最小代价",
            "translatedContent": "<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;。</p>\n\n<p>你可以对 <code>nums</code>&nbsp;执行特殊操作 <strong>任意次</strong>&nbsp;（也可以 <strong>0</strong>&nbsp;次）。每一次特殊操作中，你需要 <strong>按顺序</strong>&nbsp;执行以下步骤：</p>\n\n<ul>\n\t<li>从范围&nbsp;<code>[0, n - 1]</code>&nbsp;里选择一个下标 <code>i</code>&nbsp;和一个 <strong>正</strong>&nbsp;整数&nbsp;<code>x</code>&nbsp;。</li>\n\t<li>将&nbsp;<code>|nums[i] - x|</code>&nbsp;添加到总代价里。</li>\n\t<li>将 <code>nums[i]</code>&nbsp;变为&nbsp;<code>x</code>&nbsp;。</li>\n</ul>\n\n<p>如果一个正整数正着读和反着读都相同，那么我们称这个数是<strong>&nbsp;回文数</strong>&nbsp;。比方说，<code>121</code>&nbsp;，<code>2552</code> 和&nbsp;<code>65756</code>&nbsp;都是回文数，但是&nbsp;<code>24</code>&nbsp;，<code>46</code>&nbsp;，<code>235</code>&nbsp;都不是回文数。</p>\n\n<p>如果一个数组中的所有元素都等于一个整数&nbsp;<code>y</code>&nbsp;，且&nbsp;<code>y</code>&nbsp;是一个小于&nbsp;<code>10<sup>9</sup></code>&nbsp;的&nbsp;<strong>回文数</strong>&nbsp;，那么我们称这个数组是一个 <strong>等数数组&nbsp;</strong>。</p>\n\n<p>请你返回一个整数，表示执行任意次特殊操作后使 <code>nums</code>&nbsp;成为 <strong>等数数组</strong>&nbsp;的 <strong>最小</strong>&nbsp;总代价。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,2,3,4,5]\n<b>输出：</b>6\n<b>解释：</b>我们可以将数组中所有元素变为回文数 3 得到等数数组，数组变成 [3,3,3,3,3] 需要执行 4 次特殊操作，代价为 |1 - 3| + |2 - 3| + |4 - 3| + |5 - 3| = 6 。\n将所有元素变为其他回文数的总代价都大于 6 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [10,12,13,14,15]\n<b>输出：</b>11\n<b>解释：</b>我们可以将数组中所有元素变为回文数 11 得到等数数组，数组变成 [11,11,11,11,11] 需要执行 5 次特殊操作，代价为 |10 - 11| + |12 - 11| + |13 - 11| + |14 - 11| + |15 - 11| = 11 。\n将所有元素变为其他回文数的总代价都大于 11 。\n</pre>\n\n<p><strong class=\"example\">示例 3 ：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [22,33,22,33,22]\n<b>输出：</b>22\n<b>解释：</b>我们可以将数组中所有元素变为回文数 22 得到等数数组，数组变为 [22,22,22,22,22] 需要执行 2 次特殊操作，代价为 |33 - 22| + |33 - 22| = 22 。\n将所有元素变为其他回文数的总代价都大于 22 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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                "Find the median of <code>nums</code> after sorting it (if the length is even, we can select any number from the two in the middle). Let’s call it <code>m</code>.",
                "Try the smallest palindromic number that is larger than or equal to <code>m</code> (if any) and the largest palindromic number that is smaller than or equal to <code>m</code> (if any). These two values are the candidate palindromic numbers for values of all indices.",
                "We can use math constructions to construct the two palindromic numbers in <code>O(log(m) / 2)</code> time or we can do it using brute-force by starting from m and checking smaller and larger values in <code>O(sqrt(10<sup>log(m)</sup>))</code>.",
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