leetcode-problemset/leetcode-cn/originData/find-the-k-or-of-an-array.json

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        "question": {
            "questionId": "3183",
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            "categoryTitle": "Algorithms",
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            "title": "Find the K-or of an Array",
            "titleSlug": "find-the-k-or-of-an-array",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and an integer <code>k</code>.</p>\n\n<p>The <strong>K-or</strong> of <code>nums</code> is a non-negative integer that satisfies the following:</p>\n\n<ul>\n\t<li>The <code>i<sup>th</sup></code> bit is set in the K-or <strong>if and only if</strong> there are at least <code>k</code> elements of nums in which bit <code>i</code> is set.</li>\n</ul>\n\n<p>Return <em>the <strong> K-or</strong> of</em> <code>nums</code>.</p>\n\n<p><strong>Note</strong> that a bit <code>i</code> is set in <code>x</code> if <code>(2<sup>i</sup> AND x) == 2<sup>i</sup></code>, where <code>AND</code> is the bitwise <code>AND</code> operator.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [7,12,9,8,9,15], k = 4\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> Bit 0 is set at nums[0], nums[2], nums[4], and nums[5].\nBit 1 is set at nums[0], and nums[5].\nBit 2 is set at nums[0], nums[1], and nums[5].\nBit 3 is set at nums[1], nums[2], nums[3], nums[4], and nums[5].\nOnly bits 0 and 3 are set in at least k elements of the array, and bits i &gt;= 4 are not set in any of the array&#39;s elements. Hence, the answer is 2^0 + 2^3 = 9.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,12,1,11,4,5], k = 6\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Since k == 6 == nums.length, the 6-or of the array is equal to the bitwise AND of all its elements. Hence, the answer is 2 AND 12 AND 1 AND 11 AND 4 AND 5 = 0.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [10,8,5,9,11,6,8], k = 1\n<strong>Output:</strong> 15\n<strong>Explanation:</strong> Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.\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;= 50</code></li>\n\t<li><code>0 &lt;= nums[i] &lt; 2<sup>31</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= nums.length</code></li>\n</ul>\n",
            "translatedTitle": "找出数组中的 K-or 值",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 和一个整数 <code>k</code> 。</p>\n\n<p><code>nums</code> 中的 <strong>K-or</strong> 是一个满足以下条件的非负整数：</p>\n\n<ul>\n\t<li>只有在 <code>nums</code> 中，至少存在 <code>k</code> 个元素的第 <code>i</code> 位值为 1 ，那么 K-or 中的第 <code>i</code> 位的值才是 1 。</li>\n</ul>\n\n<p>返回 <code>nums</code> 的 <strong>K-or</strong> 值。</p>\n\n<p><strong>注意</strong> ：对于整数 <code>x</code> ，如果&nbsp;<code>(2<sup>i</sup> AND x) == 2<sup>i</sup></code> ，则 <code>x</code> 中的第 <code>i</code> 位值为 1 ，其中 <code>AND</code> 为按位与运算符。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [7,12,9,8,9,15], k = 4\n<strong>输出：</strong>9\n<strong>解释：</strong>nums[0]、nums[2]、nums[4] 和 nums[5] 的第 0 位的值为 1 。\nnums[0] 和 nums[5] 的第 1 位的值为 1 。\nnums[0]、nums[1] 和 nums[5] 的第 2 位的值为 1 。\nnums[1]、nums[2]、nums[3]、nums[4] 和 nums[5] 的第 3 位的值为 1 。\n只有第 0 位和第 3 位满足数组中至少存在 k 个元素在对应位上的值为 1 。因此，答案为 2^0 + 2^3 = 9 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,12,1,11,4,5], k = 6\n<strong>输出：</strong>0\n<strong>解释：</strong>因为 k == 6 == nums.length ，所以数组的 6-or 等于其中所有元素按位与运算的结果。因此，答案为 2 AND 12 AND 1 AND 11 AND 4 AND 5 = 0 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [10,8,5,9,11,6,8], k = 1\n<strong>输出：</strong>15\n<strong>解释：</strong>因为 k == 1 ，数组的 1-or 等于其中所有元素按位或运算的结果。因此，答案为 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15 。</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;= 50</code></li>\n\t<li><code>0 &lt;= nums[i] &lt; 2<sup>31</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= nums.length</code></li>\n</ul>\n",
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