We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Divide an Array Into Subarrays With Minimum Cost II

Difficulty: Hard

Problem Description

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. 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. Return the minimum possible sum of the cost of these subarrays.

Key Insights

  • The first element of each subarray determines its cost.
  • The placement of subarray boundaries is constrained by the distance dist between the first and last subarray starts.
  • A dynamic programming approach can help minimize the total cost by storing intermediate results.
  • Using a priority queue can efficiently help track the minimum costs as we explore different partitions.

Space and Time Complexity

Time Complexity: O(n log k)
Space Complexity: O(k)

Solution

We can approach this problem using dynamic programming combined with a priority queue (min-heap). The idea is to maintain a dp array where dp[i] represents the minimum cost to divide the first i elements into a valid number of subarrays. We iterate through the array, and for each possible endpoint of a subarray, we check valid starting points within the bounds defined by dist. The priority queue helps efficiently retrieve the minimum cost for the previous subarray partitions.

Code Solutions

import heapq

def minCostSubarrays(nums, k, dist):
    n = len(nums)
    dp = [float('inf')] * n
    dp[0] = nums[0]  # Cost for the first subarray

    for i in range(1, n):
        # For each end of the current subarray
        min_cost = float('inf')
        for j in range(max(0, i - dist), i):
            min_cost = min(min_cost, dp[j])
        
        dp[i] = min_cost + nums[i]  # Update dp with the new cost

    return sum(dp[i] for i in range(k-1, n))  # Return the sum of the k minimum costs
← Back to All Questions