You are given a 0-indexed integer array nums and an integer k.
In one operation, you can pick any index i of nums such that 0 <= i < nums.length - 1 and replace nums[i] and nums[i + 1] with a single occurrence of nums[i] & nums[i + 1], where & represents the bitwise AND operator.
Return the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.
\n
Example 1:
\n\n\nInput: nums = [3,5,3,2,7], k = 2\nOutput: 3\nExplanation: Let's do the following operations:\n1. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [1,3,2,7].\n2. Replace nums[2] and nums[3] with (nums[2] & 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.\n\n
Example 2:
\n\n\nInput: nums = [7,3,15,14,2,8], k = 4\nOutput: 2\nExplanation: Let's do the following operations:\n1. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,15,14,2,8]. \n2. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,14,2,8].\n3. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [2,2,8].\n4. Replace nums[1] and nums[2] with (nums[1] & 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\n\n
Example 3:
\n\n\nInput: nums = [10,7,10,3,9,14,9,4], k = 1\nOutput: 15\nExplanation: 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\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1050 <= nums[i] < 2300 <= k < nums.length给你一个下标从 0 开始的整数数组 nums 和一个整数 k 。
一次操作中，你可以选择 nums 中满足 0 <= i < nums.length - 1 的一个下标 i ，并将 nums[i] 和 nums[i + 1] 替换为数字 nums[i] & nums[i + 1] ，其中 & 表示按位 AND 操作。
请你返回 至多 k 次操作以内，使 nums 中所有剩余元素按位 OR 结果的 最小值 。
\n\n
示例 1：
\n\n\n输入：nums = [3,5,3,2,7], k = 2\n输出：3\n解释：执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [1,3,2,7] 。\n2. 将 nums[2] 和 nums[3] 替换为 (nums[2] & 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。\n\n
示例 2：
\n\n\n输入：nums = [7,3,15,14,2,8], k = 4\n输出：2\n解释：执行以下操作：\n1. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [3,15,14,2,8] 。\n2. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [3,14,2,8] 。\n3. 将 nums[0] 和 nums[1] 替换为 (nums[0] & nums[1]) ，得到 nums 为 [2,2,8] 。\n4. 将 nums[1] 和 nums[2] 替换为 (nums[1] & 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\n\n
示例 3：
\n\n\n输入：nums = [10,7,10,3,9,14,9,4], k = 1\n输出：15\n解释：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\n\n
\n\n
提示：
\n\n1 <= nums.length <= 1050 <= nums[i] < 2300 <= k < nums.lengthmask.",
"For a fixed bit, add it to mask then check if there exists some sequence of k operations such that mask & answer == 0 where answer is the bitwise-or of the remaining elements of nums. If there is no such sequence of operations, remove the current bit from mask. How can we perform this check?",
"Let x be the bitwise-and of all elements of nums. If x AND mask != 0, there is no sequence of operations that satisfies the condition in the previous hint. This is because even if we perform this operation n - 1 times on the array, we will end up with x 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 k. Let’s calculate the minimum number of operations in such a sequence.",
"Iterate over the array from left to right, if nums[i] & mask != 0, apply the operation on index i.",
"After iterating over all elements, let x be the bitwise-and of all elements of nums. If x == 0, then we have found the minimum number of operations. Otherwise, It can be proven that we need exactly one more operation so that x == 0.",
"The condition in the second hint is satisfied if and only if the minimum number of operations is less than or equal to k."
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