Introduction

In this blog post, we will explore how to solve the LeetCode problem "3013" 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 a 0-indexed array of integers nums of length n, and two positive integers k and dist. The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3. You need to divide nums into k disjoint contiguous subarrays, such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist. In other words, if you divide nums into the subarrays nums[0..(i1 - 1)], nums[i1..(i2 - 1)], ..., nums[ik-1..(n - 1)], then ik-1 - i1 <= dist. Return the minimum possible sum of the cost of these subarrays. Example 1: Input: nums = [1,3,2,6,4,2], k = 3, dist = 3 Output: 5 Explanation: The best possible way to divide nums into 3 subarrays is: [1,3], [2,6,4], and [2]. This choice is valid because ik-1 - i1 is 5 - 2 = 3 which is equal to dist. The total cost is nums[0] + nums[2] + nums[5] which is 1 + 2 + 2 = 5. It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 5. Example 2: Input: nums = [10,1,2,2,2,1], k = 4, dist = 3 Output: 15 Explanation: The best possible way to divide nums into 4 subarrays is: [10], [1], [2], and [2,2,1]. This choice is valid because ik-1 - i1 is 3 - 1 = 2 which is less than dist. The total cost is nums[0] + nums[1] + nums[2] + nums[3] which is 10 + 1 + 2 + 2 = 15. The division [10], [1], [2,2,2], and [1] is not valid, because the difference between ik-1 and i1 is 5 - 1 = 4, which is greater than dist. It can be shown that there is no possible way to divide nums into 4 subarrays at a cost lower than 15. Example 3: Input: nums = [10,8,18,9], k = 3, dist = 1 Output: 36 Explanation: The best possible way to divide nums into 4 subarrays is: [10], [8], and [18,9]. This choice is valid because ik-1 - i1 is 2 - 1 = 1 which is equal to dist.The total cost is nums[0] + nums[1] + nums[2] which is 10 + 8 + 18 = 36. The division [10], [8,18], and [9] is not valid, because the difference between ik-1 and i1 is 3 - 1 = 2, which is greater than dist. It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 36. Constraints: 3 <= n <= 105 1 <= nums[i] <= 109 3 <= k <= n k - 2 <= dist <= n - 2

Explanation

Here's the breakdown of the approach, complexity, and the Python code solution:

• High-Level Approach:

  • Dynamic Programming: Use DP to store minimum costs for dividing the array into i subarrays ending at index j.
  • Iteration and Optimization: Iterate through possible starting positions of the last subarray while enforcing the dist constraint.
  • Base Case: The cost of one subarray starting at index 0 is nums[0].

• Complexity:

  • Runtime: O(n * dist)
  • Storage: O(n * k)

Code

    def minCost(nums, k, dist):
    n = len(nums)
    dp = [[float('inf')] * n for _ in range(k)]

    # Base case: cost of one subarray starting at index 0
    for j in range(n):
        dp[0][j] = nums[0]

    # Iterate to build up DP table
    for i in range(1, k):
        for j in range(i, n):  # j must be at least i
            for l in range(max(i - 1, 0), j):
                if i == k -1:
                    if j - (l+1) <= dist:
                        dp[i][j] = min(dp[i][j], dp[i - 1][l] + nums[l + 1])
                else:
                    dp[i][j] = min(dp[i][j], dp[i - 1][l] + nums[l + 1])

    return dp[k - 1][n - 1]