leetcode-problemset/leetcode-cn/originData/minimize-or-of-remaining-elements-using-operations.json

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        "question": {
            "questionId": "3261",
            "questionFrontendId": "3022",
            "categoryTitle": "Algorithms",
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            "title": "Minimize OR of Remaining Elements Using Operations",
            "titleSlug": "minimize-or-of-remaining-elements-using-operations",
            "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>In one operation, you can pick any index <code>i</code> of <code>nums</code> such that <code>0 &lt;= i &lt; nums.length - 1</code> and replace <code>nums[i]</code> and <code>nums[i + 1]</code> with a single occurrence of <code>nums[i] &amp; nums[i + 1]</code>, where <code>&amp;</code> represents the bitwise <code>AND</code> operator.</p>\n\n<p>Return <em>the <strong>minimum</strong> possible value of the bitwise </em><code>OR</code><em> of the remaining elements of</em> <code>nums</code> <em>after applying <strong>at most</strong></em> <code>k</code> <em>operations</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,5,3,2,7], k = 2\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> Let&#39;s do the following operations:\n1. Replace nums[0] and nums[1] with (nums[0] &amp; nums[1]) so that nums becomes equal to [1,3,2,7].\n2. Replace nums[2] and nums[3] with (nums[2] &amp; nums[3]) so that nums becomes equal to [1,3,2].\nThe bitwise-or of the final array is 3.\nIt can be shown that 3 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [7,3,15,14,2,8], k = 4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Let&#39;s do the following operations:\n1. Replace nums[0] and nums[1] with (nums[0] &amp; nums[1]) so that nums becomes equal to [3,15,14,2,8]. \n2. Replace nums[0] and nums[1] with (nums[0] &amp; nums[1]) so that nums becomes equal to [3,14,2,8].\n3. Replace nums[0] and nums[1] with (nums[0] &amp; nums[1]) so that nums becomes equal to [2,2,8].\n4. Replace nums[1] and nums[2] with (nums[1] &amp; nums[2]) so that nums becomes equal to [2,0].\nThe bitwise-or of the final array is 2.\nIt can be shown that 2 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [10,7,10,3,9,14,9,4], k = 1\n<strong>Output:</strong> 15\n<strong>Explanation:</strong> Without applying any operations, the bitwise-or of nums is 15.\nIt can be shown that 15 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.\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;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt; 2<sup>30</sup></code></li>\n\t<li><code>0 &lt;= k &lt; nums.length</code></li>\n</ul>\n",
            "translatedTitle": "给定操作次数内使剩余元素的或值最小",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;。</p>\n\n<p>一次操作中，你可以选择 <code>nums</code>&nbsp;中满足&nbsp;<code>0 &lt;= i &lt; nums.length - 1</code>&nbsp;的一个下标 <code>i</code>&nbsp;，并将&nbsp;<code>nums[i]</code> 和&nbsp;<code>nums[i + 1]</code>&nbsp;替换为数字&nbsp;<code>nums[i] &amp; nums[i + 1]</code>&nbsp;，其中&nbsp;<code>&amp;</code>&nbsp;表示按位&nbsp;<code>AND</code>&nbsp;操作。</p>\n\n<p>请你返回 <strong>至多</strong>&nbsp;<code>k</code>&nbsp;次操作以内，使 <code>nums</code>&nbsp;中所有剩余元素按位 <code>OR</code>&nbsp;结果的 <strong>最小值</strong>&nbsp;。</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 = [3,5,3,2,7], k = 2\n<b>输出：</b>3\n<b>解释：</b>执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] &amp; nums[1]) ，得到 nums 为 [1,3,2,7] 。\n2. 将 nums[2] 和 nums[3] 替换为 (nums[2] &amp; nums[3]) ，得到 nums 为 [1,3,2] 。\n最终数组的按位或值为 3 。\n3 是 k 次操作以内，可以得到的剩余元素的最小按位或值。</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [7,3,15,14,2,8], k = 4\n<b>输出：</b>2\n<b>解释：</b>执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] &amp; nums[1]) ，得到 nums 为 [3,15,14,2,8] 。\n2. 将 nums[0] 和 nums[1] 替换为 (nums[0] &amp; nums[1]) ，得到 nums 为 [3,14,2,8] 。\n3. 将 nums[0] 和 nums[1] 替换为 (nums[0] &amp; nums[1]) ，得到 nums 为 [2,2,8] 。\n4. 将 nums[1] 和 nums[2] 替换为 (nums[1] &amp; nums[2]) ，得到 nums 为 [2,0] 。\n最终数组的按位或值为 2 。\n2 是 k 次操作以内，可以得到的剩余元素的最小按位或值。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [10,7,10,3,9,14,9,4], k = 1\n<b>输出：</b>15\n<b>解释：</b>不执行任何操作，nums 的按位或值为 15 。\n15 是 k 次操作以内，可以得到的剩余元素的最小按位或值。\n</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;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt; 2<sup>30</sup></code></li>\n\t<li><code>0 &lt;= k &lt; nums.length</code></li>\n</ul>\n",
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            "hints": [
                "From the most significant bit to the least significant bit, maintain the bits that will not be included in the final answer in a variable <code>mask</code>.",
                "For a fixed bit, add it to <code>mask</code> then check if there exists some sequence of <code>k</code> operations such that <code>mask & answer == 0 </code> where <code>answer</code> is the bitwise-or of the remaining elements of <code>nums</code>. If there is no such sequence of operations, remove the current bit from <code>mask</code>. How can we perform this check?",
                "Let <code>x</code> be the bitwise-and of all elements of <code>nums</code>. If <code>x AND mask != 0</code>, there is no sequence of operations that satisfies the condition in the previous hint. This is because even if we perform this operation <code>n - 1</code> times on the array, we will end up with <code>x</code> as the final element.",
                "Otherwise, there exists at least one such sequence. It is sufficient to check if the number of operations in such a sequence is less than <code>k</code>. Let’s calculate the minimum number of operations in such a sequence.",
                "Iterate over the array from left to right, if <code>nums[i] & mask != 0</code>, apply the operation on index <code>i</code>.",
                "After iterating over all elements, let <code>x</code> be the bitwise-and of all elements of <code>nums</code>. If <code>x == 0</code>, then we have found the minimum number of operations. Otherwise, It can be proven that we need exactly one more operation so that <code>x == 0</code>.",
                "The condition in the second hint is satisfied if and only if the minimum number of operations is less than or equal to <code>k</code>."
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