Introduction

In this blog post, we will explore how to solve the LeetCode problem "3117" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 <= 104 1 <= m == andValues.length <= min(n, 10) 1 <= nums[i] < 105 0 <= andValues[j] < 105

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming to explore all possible ways to divide the nums array into m subarrays. dp[i][j] stores the minimum sum of values for dividing the first i elements of nums into j subarrays, satisfying the AND conditions.

• Iterative Calculation: The dp table is built iteratively. For each dp[i][j], we consider all possible split points k (from j-1 to i-1). We calculate the bitwise AND of the subarray nums[k:i] and check if it matches andValues[j-1]. If it does, we update dp[i][j] with the minimum of its current value and dp[k][j-1] + nums[i-1].

• Base Cases and Result: The base case is dp[0][0] = 0. If dp[n][m] remains infinity after the calculations, it means no valid division is possible, and we return -1. Otherwise, we return dp[n][m].

• Runtime Complexity: O(n² * m), where n is the length of nums and m is the length of andValues. The nested loops dominate the runtime.

• Storage Complexity: O(n * m) due to the dp table.

Code

    def min_subarray_value_sum(nums, andValues):
    n = len(nums)
    m = len(andValues)

    if m > n:
        return -1

    dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(1, n + 1):
        for j in range(1, m + 1):
            for k in range(j - 1, i):
                current_and = nums[k]
                for l in range(k + 1, i):
                    current_and &= nums[l]

                if current_and == andValues[j - 1] and dp[k][j - 1] != float('inf'):
                    dp[i][j] = min(dp[i][j], dp[k][j - 1] + nums[i - 1])

    if dp[n][m] == float('inf'):
        return -1
    else:
        return dp[n][m]