leetcode-problemset/leetcode-cn/originData/minimum-array-end.json

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        "question": {
            "questionId": "3394",
            "questionFrontendId": "3133",
            "categoryTitle": "Algorithms",
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            "title": "Minimum Array End",
            "titleSlug": "minimum-array-end",
            "content": "<p>You are given two integers <code>n</code> and <code>x</code>. You have to construct an array of <strong>positive</strong> integers <code>nums</code> of size <code>n</code> where for every <code>0 &lt;= i &lt; n - 1</code>, <code>nums[i + 1]</code> is <strong>greater than</strong> <code>nums[i]</code>, and the result of the bitwise <code>AND</code> operation between all elements of <code>nums</code> is <code>x</code>.</p>\n\n<p>Return the <strong>minimum</strong> possible value of <code>nums[n - 1]</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 3, x = 4</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">6</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p><code>nums</code> can be <code>[4,5,6]</code> and its last element is 6.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 2, x = 7</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">15</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p><code>nums</code> can be <code>[7,15]</code> and its last element is 15.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n, x &lt;= 10<sup>8</sup></code></li>\n</ul>\n",
            "translatedTitle": "数组最后一个元素的最小值",
            "translatedContent": "<p>给你两个整数 <code>n</code> 和 <code>x</code> 。你需要构造一个长度为 <code>n</code> 的 <strong>正整数 </strong>数组 <code>nums</code> ，对于所有 <code>0 &lt;= i &lt; n - 1</code> ，满足 <code>nums[i + 1]</code><strong> 大于 </strong><code>nums[i]</code> ，并且数组 <code>nums</code> 中所有元素的按位 <code>AND</code> 运算结果为 <code>x</code> 。</p>\n\n<p>返回 <code>nums[n - 1]</code> 可能的<strong> 最小 </strong>值。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">n = 3, x = 4</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">6</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>数组 <code>nums</code> 可以是 <code>[4,5,6]</code> ，最后一个元素为 <code>6</code> 。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">n = 2, x = 7</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">15</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>数组 <code>nums</code> 可以是 <code>[7,15]</code> ，最后一个元素为 <code>15</code> 。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n, x &lt;= 10<sup>8</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    long long minEnd(int n, int x) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def minEnd(self, n, x):\n        \"\"\"\n        :type n: int\n        :type x: int\n        :rtype: int\n        \"\"\"",
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                    "code": "# @param {Integer} n\n# @param {Integer} x\n# @return {Integer}\ndef min_end(n, x)\n\nend",
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            "hints": [
                "Each element of the array should be obtained by “merging” <code>x</code> and <code>v</code> where <code>v = 0, 1, 2, …(n - 1)</code>.",
                "To merge <code>x</code> with another number <code>v</code>, keep the set bits of <code>x</code> untouched, for all the other bits, fill the set bits of <code>v</code> from right to left in order one by one.",
                "So the final answer is the “merge” of <code>x</code> and <code>n - 1</code>."
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