# Solving Leetcode Interviews in Seconds with AI: Minimum Cost to Divide Array Into Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3500" 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 integer arrays, nums and cost, of the same size, and an integer k. You can divide nums into subarrays. The cost of the ith subarray consisting of elements nums[l..r] is:  (nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r]).  Note that i represents the order of the subarray: 1 for the first subarray, 2 for the second, and so on. Return the minimum total cost possible from any valid division.   Example 1:  Input: nums = [3,1,4], cost = [4,6,6], k = 1 Output: 110 Explanation: The minimum total cost possible can be achieved by dividing nums into subarrays [3, 1] and [4].   The cost of the first subarray [3,1] is (3 + 1 + 1 * 1) * (4 + 6) = 50. The cost of the second subarray [4] is (3 + 1 + 4 + 1 * 2) * 6 = 60.   Example 2:  Input: nums = [4,8,5,1,14,2,2,12,1], cost = [7,2,8,4,2,2,1,1,2], k = 7 Output: 985 Explanation: The minimum total cost possible can be achieved by dividing nums into subarrays [4, 8, 5, 1], [14, 2, 2], and [12, 1].   The cost of the first subarray [4, 8, 5, 1] is (4 + 8 + 5 + 1 + 7 * 1) * (7 + 2 + 8 + 4) = 525. The cost of the second subarray [14, 2, 2] is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 7 * 2) * (2 + 2 + 1) = 250. The cost of the third subarray [12, 1] is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 12 + 1 + 7 * 3) * (1 + 2) = 210.     Constraints:  1 <= nums.length <= 1000 cost.length == nums.length 1 <= nums[i], cost[i] <= 1000 1 <= k <= 1000  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming:** Use dynamic programming to build up the minimum cost by considering all possible subarray divisions. `dp[i]` stores the minimum cost to divide the first `i` elements.

*   **Iterative Calculation:** Iterate through all possible ending points `i` of a subarray and, for each `i`, iterate through all possible starting points `j` of the subarray (from 0 to i). Calculate the cost of this subarray and update `dp[i]` with the minimum cost found so far.

*   **Prefix Sums:** Use prefix sums for `nums` and `cost` arrays to calculate subarray sums efficiently.

*   **Time Complexity:** O(n^2), **Space Complexity:** O(n)

	
	# Code
	```python
	def min_total_cost(nums, cost, k):
    n = len(nums)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0

    prefix_nums = [0] * (n + 1)
    prefix_cost = [0] * (n + 1)

    for i in range(1, n + 1):
        prefix_nums[i] = prefix_nums[i - 1] + nums[i - 1]
        prefix_cost[i] = prefix_cost[i - 1] + cost[i - 1]

    for i in range(1, n + 1):
        for j in range(i):
            subarray_nums_sum = prefix_nums[i] - prefix_nums[j]
            subarray_cost_sum = prefix_cost[i] - prefix_cost[j]
            
            current_cost = (prefix_nums[i] + k * (dp[j] == float('inf') or (dp[j] > 0)) ) * subarray_cost_sum
            if dp[j] == float('inf'):
              current_cost = (subarray_nums_sum + k * 1) * subarray_cost_sum
            else:
              sub_count = 0
              temp = dp[j]
              if temp > 0:
                  sub_count = 1
              
              l = 0
              count = 1
              while l < j:
                  l +=1
                  
              tempdp = [float('inf')] * (j + 1)
              tempdp[0] = 0
              for x in range(1, j + 1):
                  for y in range(x):
                      
                      sub_nums_sum = prefix_nums[x] - prefix_nums[y]
                      sub_cost_sum = prefix_cost[x] - prefix_cost[y]
                      
                      if tempdp[y] == float('inf'):
                          continue
                      
                      
                      tempdp[x] = min(tempdp[x], (sub_nums_sum + (prefix_nums[y] if y > 0 else 0) + k * (tempdp[y] == float('inf') or tempdp[y] > 0)) * sub_cost_sum + (tempdp[y] if tempdp[y] != float('inf') else 0) )
                      

              count = 0

              y = 0
              while y < j:
                  
                  sub_nums_sum = prefix_nums[j] - prefix_nums[y]
                  sub_cost_sum = prefix_cost[j] - prefix_cost[y]
                  if tempdp[y] == float('inf'):
                      continue
                  count+=1

            
              if dp[j] == float('inf'):
                current_cost = (subarray_nums_sum + k * 1) * subarray_cost_sum
              else:
                current_cost = (subarray_nums_sum + prefix_nums[j] + k * (count + 1) ) * subarray_cost_sum + dp[j]

            dp[i] = min(dp[i], current_cost if dp[j] != float('inf') else (subarray_nums_sum + k * 1) * subarray_cost_sum)

    return dp[n]
	```