3117. Minimum Sum of Values by Dividing Array

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def minimumValueSum(self, nums: List[int], andValues: List[int]) -> int:
        N = len(nums)
        M = len(andValues)
        defaultMask = (1 << 18) - 1
 
        @cache
        def go(i, k, mask):
            if k == M and i == N: return 0
            if k == M or i == N: return inf
 
            mask &= nums[i]
            if mask < andValues[k]: return inf
            if mask == andValues[k]: return min(go(i + 1, k, mask), nums[i] + go(i + 1, k + 1, defaultMask))
            if mask > andValues[k]: return go(i + 1, k, mask)
        
        ans = go(0, 0, defaultMask)
        return -1 if ans == inf else ans