3117 - Minimum Sum of Values by Dividing Array.

Hard

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 <code>i^th</code> subarray <code>l_i, r_i</code>, the bitwise AND of the subarray elements is equal to andValues[i], in other words, <code>numsl_i& numsl_i + 1& ... & numsr_i == andValuesi</code> 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:

• <code>1 <= n == nums.length <= 10⁴</code>

• 1 <= m == andValues.length <= min(n, 10)

• <code>1 <= numsi< 10⁵</code>

• <code>0 <= andValuesj< 10⁵</code>