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.
You 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.
Example 1:
Input: nums = [1,4,3,3,2], andValues = [0,3,3,2]
Output: 12
Explanation:
The 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:
Input: nums = [2,3,5,7,7,7,5], andValues = [0,7,5]
Output: 17
Explanation:
There 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:
Input: nums = [1,2,3,4], andValues = [2]
Output: -1
Explanation:
The 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.
Constraints:
1 <= n == nums.length <= 1041 <= m == andValues.length <= min(n, 10)1 <= nums[i] < 1050 <= andValues[j] < 105