What is the brute force solution for Divide an Array Into Subarrays With Minimum Cost II?

The brute force approach for Divide an Array Into Subarrays With Minimum Cost II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to divide the array into k subarrays and calculates the cost for each division. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Divide an Array Into Subarrays With Minimum Cost II?

Divide an Array Into Subarrays With Minimum Cost II uses the following concepts: Array, Hash Table, Sliding Window, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Divide an Array Into Subarrays With Minimum Cost II?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and potential optimizations as you work through the problem. Practice similar problems to get comfortable with dynamic programming and heap usage.

What are common mistakes when solving Divide an Array Into Subarrays With Minimum Cost II?

Failing to check the distance condition properly when generating combinations. Not using efficient data structures like heaps to manage costs dynamically.

#3013

Divide an Array Into Subarrays With Minimum Cost II

Hard

ArrayHash TableSliding WindowHeap (Priority Queue)Hash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a dynamic programming approach combined with a max heap to efficiently track the smallest costs while adhering to the distance constraint. This reduces the number of combinations we need to check.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[j] represents the minimum cost to divide the first i elements into j subarrays.
2Step 2: Use a max heap to keep track of the smallest costs dynamically as we iterate through the array.
3Step 3: For each starting index of a subarray, calculate the cost and update the DP array based on previous results and the heap.

solution.py19 lines

1# Full working Python code
2import heapq
3
4def minCost(nums, k, dist):
5    n = len(nums)
6    dp = [float('inf')] * (k + 1)
7    dp[0] = 0
8    for i in range(n):
9        new_dp = dp[:]
10        max_heap = []
11        for j in range(i + 1):
12            if j > 0:
13                heapq.heappush(max_heap, -nums[j - 1])
14            while max_heap and j - i > dist:
15                heapq.heappop(max_heap)
16            if len(max_heap) >= k - 1:
17                new_dp[k] = min(new_dp[k], dp[k - 1] - max_heap[0] + nums[i])
18        dp = new_dp
19    return dp[k]

ℹ

Complexity note: The time complexity is reduced to O(n log n) due to the use of a max heap for maintaining the smallest costs dynamically, allowing us to efficiently update our DP array.

1Understanding the cost structure is crucial; it always depends on the first element of each subarray.
2The distance constraint limits the possible combinations, which can be leveraged to optimize the solution.

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