You are given an array nums and an integer k. You need to find a subarray of nums such that the absolute difference between k and the bitwise AND of the subarray elements is as small as possible. In other words, select a subarray nums[l..r] such that |k - (nums[l] AND nums[l + 1] ... AND nums[r])| is minimum.
Return the minimum possible value of the absolute difference.
\n\nA subarray is a contiguous non-empty sequence of elements within an array.
\n\n\n
Example 1:
\n\nInput: nums = [1,2,4,5], k = 3
\n\nOutput: 1
\n\nExplanation:
\n\nThe subarray nums[2..3] has AND value 4, which gives the minimum absolute difference |3 - 4| = 1.
Example 2:
\n\nInput: nums = [1,2,1,2], k = 2
\n\nOutput: 0
\n\nExplanation:
\n\nThe subarray nums[1..1] has AND value 2, which gives the minimum absolute difference |2 - 2| = 0.
Example 3:
\n\nInput: nums = [1], k = 10
\n\nOutput: 9
\n\nExplanation:
\n\nThere is a single subarray with AND value 1, which gives the minimum absolute difference |10 - 1| = 9.
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 109给你一个数组 nums 和一个整数 k 。你需要找到 nums 的一个 子数组 ，满足子数组中所有元素按位与运算 AND 的值与 k 的 绝对差 尽可能 小 。换言之，你需要选择一个子数组 nums[l..r] 满足 |k - (nums[l] AND nums[l + 1] ... AND nums[r])| 最小。
请你返回 最小 的绝对差值。
\n\n子数组是数组中连续的 非空 元素序列。
\n\n\n\n
示例 1：
\n\n输入：nums = [1,2,4,5], k = 3
\n\n输出：1
\n\n解释：
\n\n子数组 nums[2..3] 的按位 AND 运算值为 4 ，得到最小差值 |3 - 4| = 1 。
示例 2：
\n\n输入：nums = [1,2,1,2], k = 2
\n\n输出：0
\n\n解释：
\n\n子数组 nums[1..1] 的按位 AND 运算值为 2 ，得到最小差值 |2 - 2| = 0 。
示例 3：
\n\n输入：nums = [1], k = 10
\n\n输出：9
\n\n解释：
\n\n只有一个子数组，按位 AND 运算值为 1 ，得到最小差值 |10 - 1| = 9 。
\n\n
提示：
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 109dp[i] be the set of all the bitwise AND of all the subarrays ending at index i.",
"We start from nums[i], taking the bitwise AND result by including elements one by one from i towards left. Notice that only set bits can become unset on adding a element, and unset bits never become set again.",
"Hence dp[i] can contain at most 30 elements."
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