leetcode-problemset/leetcode-cn/originData/minimum-non-zero-product-of-the-array-elements.json

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            "title": "Minimum Non-Zero Product of the Array Elements",
            "titleSlug": "minimum-non-zero-product-of-the-array-elements",
            "content": "<p>You are given a positive integer <code>p</code>. Consider an array <code>nums</code> (<strong>1-indexed</strong>) that consists of the integers in the <strong>inclusive</strong> range <code>[1, 2<sup>p</sup> - 1]</code> in their binary representations. You are allowed to do the following operation <strong>any</strong> number of times:</p>\n\n<ul>\n\t<li>Choose two elements <code>x</code> and <code>y</code> from <code>nums</code>.</li>\n\t<li>Choose a bit in <code>x</code> and swap it with its corresponding bit in <code>y</code>. Corresponding bit refers to the bit that is in the <strong>same position</strong> in the other integer.</li>\n</ul>\n\n<p>For example, if <code>x = 11<u>0</u>1</code> and <code>y = 00<u>1</u>1</code>, after swapping the <code>2<sup>nd</sup></code> bit from the right, we have <code>x = 11<u>1</u>1</code> and <code>y = 00<u>0</u>1</code>.</p>\n\n<p>Find the <strong>minimum non-zero</strong> product of <code>nums</code> after performing the above operation <strong>any</strong> number of times. Return <em>this product</em><em> <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>.</p>\n\n<p><strong>Note:</strong> The answer should be the minimum product <strong>before</strong> the modulo operation is done.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> p = 1\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> nums = [1].\nThere is only one element, so the product equals that element.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> p = 2\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> nums = [01, 10, 11].\nAny swap would either make the product 0 or stay the same.\nThus, the array product of 1 * 2 * 3 = 6 is already minimized.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> p = 3\n<strong>Output:</strong> 1512\n<strong>Explanation:</strong> nums = [001, 010, 011, 100, 101, 110, 111]\n- In the first operation we can swap the leftmost bit of the second and fifth elements.\n    - The resulting array is [001, <u>1</u>10, 011, 100, <u>0</u>01, 110, 111].\n- In the second operation we can swap the middle bit of the third and fourth elements.\n    - The resulting array is [001, 110, 0<u>0</u>1, 1<u>1</u>0, 001, 110, 111].\nThe array product is 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512, which is the minimum possible product.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= p &lt;= 60</code></li>\n</ul>\n",
            "translatedTitle": "数组元素的最小非零乘积",
            "translatedContent": "<p>给你一个正整数&nbsp;<code>p</code>&nbsp;。你有一个下标从 <strong>1</strong>&nbsp;开始的数组&nbsp;<code>nums</code>&nbsp;，这个数组包含范围&nbsp;<code>[1, 2<sup>p</sup> - 1]</code>&nbsp;内所有整数的二进制形式（两端都 <strong>包含</strong>）。你可以进行以下操作 <strong>任意</strong>&nbsp;次：</p>\n\n<ul>\n\t<li>从 <code>nums</code>&nbsp;中选择两个元素&nbsp;<code>x</code>&nbsp;和&nbsp;<code>y</code>&nbsp; 。</li>\n\t<li>选择 <code>x</code>&nbsp;中的一位与 <code>y</code>&nbsp;对应位置的位交换。对应位置指的是两个整数 <strong>相同位置</strong>&nbsp;的二进制位。</li>\n</ul>\n\n<p>比方说，如果&nbsp;<code>x = 11<em><strong>0</strong></em>1</code>&nbsp;且&nbsp;<code>y = 00<em><strong>1</strong></em>1</code>&nbsp;，交换右边数起第 <code>2</code>&nbsp;位后，我们得到&nbsp;<code>x = 11<em><strong>1</strong></em>1</code> 和&nbsp;<code>y = 00<em><strong>0</strong></em>1</code>&nbsp;。</p>\n\n<p>请你算出进行以上操作 <strong>任意次</strong>&nbsp;以后，<code>nums</code>&nbsp;能得到的 <strong>最小非零</strong>&nbsp;乘积。将乘积对<em>&nbsp;</em><code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong> 后返回。</p>\n\n<p><strong>注意：</strong>答案应为取余 <strong>之前</strong>&nbsp;的最小值。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>p = 1\n<b>输出：</b>1\n<b>解释：</b>nums = [1] 。\n只有一个元素，所以乘积为该元素。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>p = 2\n<b>输出：</b>6\n<b>解释：</b>nums = [01, 10, 11] 。\n所有交换要么使乘积变为 0 ，要么乘积与初始乘积相同。\n所以，数组乘积 1 * 2 * 3 = 6 已经是最小值。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>p = 3\n<b>输出：</b>1512\n<b>解释：</b>nums = [001, 010, 011, 100, 101, 110, 111]\n- 第一次操作中，我们交换第二个和第五个元素最左边的数位。\n    - 结果数组为 [001, <em><strong>1</strong></em>10, 011, 100, <em><strong>0</strong></em>01, 110, 111] 。\n- 第二次操作中，我们交换第三个和第四个元素中间的数位。\n    - 结果数组为 [001, 110, 0<em><strong>0</strong></em>1, 1<em><strong>1</strong></em>0, 001, 110, 111] 。\n数组乘积 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512 是最小乘积。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= p &lt;= 60</code></li>\n</ul>\n",
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