# Solving Leetcode Interviews in Seconds with AI: Minimum Sum of Values by Dividing Array


	# 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](https://www.chatmagic.app), 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(n<sup>2</sup> * 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
	```python
	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]
	```