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

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        "question": {
            "questionId": "3261",
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            "title": "Minimize OR of Remaining Elements Using Operations",
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            "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",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int minOrAfterOperations(vector<int>& nums, int k) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public int minOrAfterOperations(int[] nums, int k) {\n        \n    }\n}",
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                    "lang": "Python",
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                    "code": "class Solution(object):\n    def minOrAfterOperations(self, nums, k):\n        \"\"\"\n        :type nums: List[int]\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public int MinOrAfterOperations(int[] nums, int k) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[]} nums\n * @param {number} k\n * @return {number}\n */\nvar minOrAfterOperations = function(nums, k) {\n    \n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $nums\n     * @param Integer $k\n     * @return Integer\n     */\n    function minOrAfterOperations($nums, $k) {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun minOrAfterOperations(nums: IntArray, k: Int): Int {\n        \n    }\n}",
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                    "code": "class Solution {\n  int minOrAfterOperations(List<int> nums, int k) {\n    \n  }\n}",
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                    "lang": "Go",
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                    "code": "func minOrAfterOperations(nums []int, k int) int {\n    \n}",
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                    "lang": "Ruby",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer} k\n# @return {Integer}\ndef min_or_after_operations(nums, k)\n    \nend",
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                    "code": "object Solution {\n    def minOrAfterOperations(nums: Array[Int], k: Int): Int = {\n        \n    }\n}",
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                    "lang": "Rust",
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                    "code": "impl Solution {\n    pub fn min_or_after_operations(nums: Vec<i32>, k: i32) -> i32 {\n        \n    }\n}",
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                    "lang": "Racket",
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                    "code": "(define/contract (min-or-after-operations nums k)\n  (-> (listof exact-integer?) exact-integer? exact-integer?)\n  )",
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                    "lang": "Erlang",
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                    "code": "-spec min_or_after_operations(Nums :: [integer()], K :: integer()) -> integer().\nmin_or_after_operations(Nums, K) ->\n  .",
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                    "code": "defmodule Solution do\n  @spec min_or_after_operations(nums :: [integer], k :: integer) :: integer\n  def min_or_after_operations(nums, k) do\n    \n  end\nend",
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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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