algoadvance

Given an integer array nums and an integer k, divide the array into subarray such that the sum of the costs of all the subarrays is minimized.

The cost of a subarray [nums_l, nums_{l+1}, …, nums_{r-1}, nums_r] is defined as:

Your task is to find the minimum cost for dividing the array into subarrays.

Example:

Input: nums = [1, 2, 3, 4, 5], k = 10
Output: 27
Explanation:
The optimal way is to divide into subarrays [1, 2, 3] and [4, 5].
The total costs are (2 * 10 + (1+2+3)) + (1 * 10 + (4+5)) = 27.

Clarifying Questions

  1. Can the array have negative integers?
  2. Is there any constraint on the size of the array?
  3. If all elements must be in separate subarrays, what should be the behavior of the function?
  4. Are there edge cases with arrays of length 1, and what should be the return value in such cases?

Strategy

  1. Dynamic Programming Approach: We will use dynamic programming to solve this problem efficiently.

  2. Subproblems Definition: Let dp[i] be the minimum cost to partition the subarray nums[:i+1].

  3. State Transition: We need to decide where to end the last subarray. Iterate j from i to 0, calculate the cost of subarray nums[j:i+1] and update dp[i] based on dp[j-1]. If j is 0, it means it’s the first subarray.

  4. Initial State: Start with dp[0] = (0 * k + nums[0]) because with one element the cost is just the single array cost as defined.

  5. Time Complexity: The solution involves nested loops, leading to a time complexity of (O(n^2)).

Code

def minCost(nums, k):
    n = len(nums)
    if n == 0:
        return 0
    dp = [float('inf')] * n
    dp[0] = 0 * k + nums[0]

    for i in range(1, n):
        current_sum = 0
        for j in range(i, -1, -1):
            current_sum += nums[j]
            if j == 0:
                subarray_cost = 0 * k + current_sum
            else:
                subarray_cost = (i - j) * k + current_sum
                dp[i] = min(dp[i], dp[j-1] + subarray_cost)

    return dp[-1]

# Example usage:
nums = [1, 2, 3, 4, 5]
k = 10
print(minCost(nums, k))  # Output: 27

Time Complexity

Try our interview co-pilot at AlgoAdvance.com