Minimum Sum of Values by Dividing Array(LeetCode 3117):

Problem

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. [1,4] as 1 & 4 == 0.
  2. [3] as the bitwise AND of a single element subarray is that element itself.
  3. [3] as the bitwise AND of a single element subarray is that element itself.
  4. [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:

  1. [[2,3,5],[7,7,7],[5]] with the sum of the values 5 + 7 + 5 == 17.
  2. [[2,3,5,7],[7,7],[5]] with the sum of the values 7 + 7 + 5 == 19.
  3. [[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 <= 104
  • 1 <= m == andValues.length <= min(n, 10)
  • 1 <= nums[i] < 105
  • 0 <= andValues[j] < 105