#3013

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Divide an Array Into Subarrays With Minimum Cost II

This problem asks to divide an array into subarrays with a minimal cost and certain constraints on subarray positions.

array hash table sliding window heap priority queue

Problem Statement

You are given an array of integers nums, with length n, and two integers k and dist. Your task is to divide the array into exactly k disjoint contiguous subarrays.

The cost of each subarray is determined by the first element of that subarray. You must ensure that the distance between the first elements of the second subarray and the k-th subarray is less than or equal to dist. Find the minimal total cost for dividing the array while respecting these constraints.

Examples

Example 1

Input: nums = [1,3,2,6,4,2], k = 3, dist = 3

Output: 5

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

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

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

Solution Approach

Dynamic Array Scanning

Begin by iterating through the array to determine possible positions for each subarray's first element. Use an array scanning technique to progressively try valid cuts while keeping track of the costs. Adjust the positions by checking the constraint on subarray distances and costs.

Hash Lookup for Optimal Cuts

Utilize a hash table to store the costs of different potential subarray partitions, allowing quick lookups of minimal costs for any given division point. This helps avoid recomputation and efficiently narrows down the optimal cost.

Sliding Window Strategy

A sliding window can be used to dynamically evaluate subarrays by adjusting their starting points and checking the distance constraint. This method allows for continuous optimization of the partitioning process with minimal recalculation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is primarily determined by the efficiency of the dynamic scanning and hash lookup operations, with potential O(n * k) complexity depending on the implementation. The space complexity depends on the storage required for dynamic partitions and hash lookups, which could be O(n) in the worst case.

What Interviewers Usually Probe

Ability to apply array scanning to partitioning problems.
Efficiency in using hash tables for dynamic lookups.
Understanding of sliding window techniques and their optimizations.

Common Pitfalls or Variants

Common pitfalls

Incorrectly dividing the array without checking the distance constraint for each subarray.
Inefficiently recalculating costs for overlapping subarrays.
Failing to optimize the partition points dynamically, leading to higher computational complexity.

Follow-up variants

Adjust the problem by increasing the number of subarrays (k) or reducing the allowed distance (dist).
Allow some overlap between subarrays to explore different partitioning strategies.
Introduce additional constraints such as limiting the maximum or minimum value of elements within subarrays.

FAQ

What is the primary approach to solving Divide an Array Into Subarrays With Minimum Cost II?

The primary approach involves dynamically scanning the array, utilizing hash lookups, and applying sliding window techniques to find the optimal division of the array into subarrays.

What are the constraints in Divide an Array Into Subarrays With Minimum Cost II?

The problem requires dividing the array into exactly k subarrays, with the distance between the first elements of the second and k-th subarrays being less than or equal to dist.

How can hash tables be used in this problem?

Hash tables help store the costs of various subarray divisions, allowing for efficient lookup and minimizing redundant computations during the partitioning process.

What is the optimal time complexity for Divide an Array Into Subarrays With Minimum Cost II?

The optimal time complexity depends on the approach, but it can range from O(n * k) to more optimized solutions depending on the exact implementation of the scanning and lookup techniques.

Can I modify the constraints of this problem to increase difficulty?

Yes, you can increase the number of subarrays or reduce the distance constraint to make the problem more complex and explore different partitioning strategies.

terminal

Solution

Solution 1: Ordered Set

The problem requires us to divide the array $\textit{nums}$ into $k$ consecutive and non-overlapping subarrays, and the distance between the first element of the second subarray and the first element of the $k$-th subarray should not exceed $\textit{dist}$. This is equivalent to finding a subarray of size $\textit{dist}+1$ starting from the element at index $1$ in $\textit{nums}$, and calculating the sum of the smallest $k-1$ elements in it. We subtract $1$ from $k$, so we only need to find the sum of the smallest $k$ elements and add $\textit{nums}[0]$ to it.

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class Solution:
    def minimumCost(self, nums: List[int], k: int, dist: int) -> int:
        def l2r():
            nonlocal s
            x = l.pop()
            s -= x
            r.add(x)

        def r2l():
            nonlocal s
            x = r.pop(0)
            l.add(x)
            s += x

        k -= 1
        s = sum(nums[: dist + 2])
        l = SortedList(nums[1 : dist + 2])
        r = SortedList()
        while len(l) > k:
            l2r()
        ans = s
        for i in range(dist + 2, len(nums)):
            x = nums[i - dist - 1]
            if x in l:
                l.remove(x)
                s -= x
            else:
                r.remove(x)
            y = nums[i]
            if y < l[-1]:
                l.add(y)
                s += y
            else:
                r.add(y)
            while len(l) < k:
                r2l()
            while len(l) > k:
                l2r()
            ans = min(ans, s)
        return ans

Continue Practicing

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