leetcode-problemset/leetcode-cn/originData/minimum-operations-to-convert-number.json

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            "title": "Minimum Operations to Convert Number",
            "titleSlug": "minimum-operations-to-convert-number",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> containing <strong>distinct</strong> numbers, an integer <code>start</code>, and an integer <code>goal</code>. There is an integer <code>x</code> that is initially set to <code>start</code>, and you want to perform operations on <code>x</code> such that it is converted to <code>goal</code>. You can perform the following operation repeatedly on the number <code>x</code>:</p>\n\n<p>If <code>0 &lt;= x &lt;= 1000</code>, then for any index <code>i</code> in the array (<code>0 &lt;= i &lt; nums.length</code>), you can set <code>x</code> to any of the following:</p>\n\n<ul>\n\t<li><code>x + nums[i]</code></li>\n\t<li><code>x - nums[i]</code></li>\n\t<li><code>x ^ nums[i]</code> (bitwise-XOR)</li>\n</ul>\n\n<p>Note that you can use each <code>nums[i]</code> any number of times in any order. Operations that set <code>x</code> to be out of the range <code>0 &lt;= x &lt;= 1000</code> are valid, but no more operations can be done afterward.</p>\n\n<p>Return <em>the <strong>minimum</strong> number of operations needed to convert </em><code>x = start</code><em> into </em><code>goal</code><em>, and </em><code>-1</code><em> if it is not possible</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,4,12], start = 2, goal = 12\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can go from 2 &rarr; 14 &rarr; 12 with the following 2 operations.\n- 2 + 12 = 14\n- 14 - 2 = 12\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,5,7], start = 0, goal = -4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can go from 0 &rarr; 3 &rarr; -4 with the following 2 operations. \n- 0 + 3 = 3\n- 3 - 7 = -4\nNote that the last operation sets x out of the range 0 &lt;= x &lt;= 1000, which is valid.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,8,16], start = 0, goal = 1\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> There is no way to convert 0 into 1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i], goal &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= start &lt;= 1000</code></li>\n\t<li><code>start != goal</code></li>\n\t<li>All the integers in <code>nums</code> are distinct.</li>\n</ul>\n",
            "translatedTitle": "转化数字的最小运算数",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，该数组由 <strong>互不相同</strong> 的数字组成。另给你两个整数 <code>start</code> 和 <code>goal</code> 。</p>\n\n<p>整数 <code>x</code> 的值最开始设为 <code>start</code> ，你打算执行一些运算使 <code>x</code> 转化为 <code>goal</code> 。你可以对数字 <code>x</code> 重复执行下述运算：</p>\n\n<p>如果 <code>0 &lt;= x &lt;= 1000</code> ，那么，对于数组中的任一下标 <code>i</code>（<code>0 &lt;= i &lt; nums.length</code>），可以将 <code>x</code> 设为下述任一值：</p>\n\n<ul>\n\t<li><code>x + nums[i]</code></li>\n\t<li><code>x - nums[i]</code></li>\n\t<li><code>x ^ nums[i]</code>（按位异或 XOR）</li>\n</ul>\n\n<p>注意，你可以按任意顺序使用每个 <code>nums[i]</code> 任意次。使 <code>x</code> 越过 <code>0 &lt;= x &lt;= 1000</code> 范围的运算同样可以生效，但该该运算执行后将不能执行其他运算。</p>\n\n<p>返回将 <code>x = start</code><em> </em>转化为<em> </em><code>goal</code><em> </em>的最小操作数；如果无法完成转化，则返回<em> </em><code>-1</code><em> </em>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,4,12], start = 2, goal = 12\n<strong>输出：</strong>2\n<strong>解释：</strong>\n可以按 2 → 14 → 12 的转化路径进行，只需执行下述 2 次运算：\n- 2 + 12 = 14\n- 14 - 2 = 12\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,5,7], start = 0, goal = -4\n<strong>输出：</strong>2\n<strong>解释：</strong>\n可以按 0 → 3 → -4 的转化路径进行，只需执行下述 2 次运算：\n- 0 + 3 = 3\n- 3 - 7 = -4\n注意，最后一步运算使 x 超过范围 0 &lt;= x &lt;= 1000 ，但该运算仍然可以生效。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,8,16], start = 0, goal = 1\n<strong>输出：</strong>-1\n<strong>解释：</strong>\n无法将 0 转化为 1</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i], goal &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= start &lt;= 1000</code></li>\n\t<li><code>start != goal</code></li>\n\t<li><code>nums</code> 中的所有整数互不相同</li>\n</ul>\n",
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