You are given two arrays nums and andValues of length n and m respectively.
The value of an array is equal to the last element of that array.
\n\nYou have to divide nums into m disjoint contiguous subarrays such that for the ith subarray [li, ri], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[li] & nums[li + 1] & ... & nums[ri] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator.
Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.
\n
Example 1:
\n\nInput: nums = [1,4,3,3,2], andValues = [0,3,3,2]
\n\nOutput: 12
\n\nExplanation:
\n\nThe only possible way to divide nums is:
[1,4] as 1 & 4 == 0.[3] as the bitwise AND of a single element subarray is that element itself.[3] as the bitwise AND of a single element subarray is that element itself.[2] as the bitwise AND of a single element subarray is that element itself.The sum of the values for these subarrays is 4 + 3 + 3 + 2 = 12.
Example 2:
\n\nInput: nums = [2,3,5,7,7,7,5], andValues = [0,7,5]
\n\nOutput: 17
\n\nExplanation:
\n\nThere are three ways to divide nums:
[[2,3,5],[7,7,7],[5]] with the sum of the values 5 + 7 + 5 == 17.[[2,3,5,7],[7,7],[5]] with the sum of the values 7 + 7 + 5 == 19.[[2,3,5,7,7],[7],[5]] with the sum of the values 7 + 7 + 5 == 19.The minimum possible sum of the values is 17.
Example 3:
\n\nInput: nums = [1,2,3,4], andValues = [2]
\n\nOutput: -1
\n\nExplanation:
\n\nThe bitwise AND of the entire array nums is 0. As there is no possible way to divide nums into a single subarray to have the bitwise AND of elements 2, return -1.
\n
Constraints:
\n\n1 <= n == nums.length <= 1041 <= m == andValues.length <= min(n, 10)1 <= nums[i] < 1050 <= andValues[j] < 105给你两个数组 nums 和 andValues，长度分别为 n 和 m。
数组的 值 等于该数组的 最后一个 元素。
\n\n你需要将 nums 划分为 m 个 不相交的连续 子数组，对于第 ith 个子数组 [li, ri]，子数组元素的按位 AND 运算结果等于 andValues[i]，换句话说，对所有的 1 <= i <= m，nums[li] & nums[li + 1] & ... & nums[ri] == andValues[i] ，其中 & 表示按位 AND 运算符。
返回将 nums 划分为 m 个子数组所能得到的可能的 最小 子数组 值 之和。如果无法完成这样的划分，则返回 -1 。
\n\n
示例 1：
\n\n输入： nums = [1,4,3,3,2], andValues = [0,3,3,2]
\n\n输出： 12
\n\n解释：
\n\n唯一可能的划分方法为：
\n\n[1,4] 因为 1 & 4 == 0[3] 因为单元素子数组的按位 AND 结果就是该元素本身[3] 因为单元素子数组的按位 AND 结果就是该元素本身[2] 因为单元素子数组的按位 AND 结果就是该元素本身这些子数组的值之和为 4 + 3 + 3 + 2 = 12
示例 2：
\n\n输入： nums = [2,3,5,7,7,7,5], andValues = [0,7,5]
\n\n输出： 17
\n\n解释：
\n\n划分 nums 的三种方式为：
[[2,3,5],[7,7,7],[5]] 其中子数组的值之和为 5 + 7 + 5 = 17[[2,3,5,7],[7,7],[5]] 其中子数组的值之和为 7 + 7 + 5 = 19[[2,3,5,7,7],[7],[5]] 其中子数组的值之和为 7 + 7 + 5 = 19子数组值之和的最小可能值为 17
示例 3：
\n\n输入： nums = [1,2,3,4], andValues = [2]
\n\n输出： -1
\n\n解释：
\n\n整个数组 nums 的按位 AND 结果为 0。由于无法将 nums 划分为单个子数组使得元素的按位 AND 结果为 2，因此返回 -1。
\n\n
提示：
\n\n1 <= n == nums.length <= 1041 <= m == andValues.length <= min(n, 10)1 <= nums[i] < 1050 <= andValues[j] < 105dp[i][j] be the optimal answer to split nums[0..(i - 1)] into the first j andValues.",
"dp[i][j] = min(dp[(i - z)][j - 1]) + nums[i - 1] over all x <= z <= y and dp[0][0] = 0, where x and y are the longest and shortest subarrays ending with nums[i - 1] and the bitwise-and of all the values in it is andValues[j - 1].",
"The answer is dp[n][m].",
"To calculate x and y, we can use binary search (or sliding window). Note that the more values we have, the smaller the AND value is.",
"To calculate the result, we need to support RMQ (range minimum query). Segment tree is one way to do it in O(log(n)). But we can use Monotonic Queue since the ranges are indeed “sliding to right” which can be reduced to the classical minimum value in sliding window problem, for a O(n) solution."
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