You are given an array of integers nums of length n.
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 3 disjoint contiguous subarrays.
Return the minimum possible sum of the cost of these subarrays.
Example 1:
Input: nums = [1,2,3,12] Output: 6 Explanation: The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6. The other possible ways to form 3 subarrays are: - [1], [2,3], and [12] at a total cost of 1 + 2 + 12 = 15. - [1,2], [3], and [12] at a total cost of 1 + 3 + 12 = 16.
Example 2:
Input: nums = [5,4,3] Output: 12 Explanation: The best possible way to form 3 subarrays is: [5], [4], and [3] at a total cost of 5 + 4 + 3 = 12. It can be shown that 12 is the minimum cost achievable.
Example 3:
Input: nums = [10,3,1,1] Output: 12 Explanation: The best possible way to form 3 subarrays is: [10,3], [1], and [1] at a total cost of 10 + 1 + 1 = 12. It can be shown that 12 is the minimum cost achievable.
Constraints:
3 <= n <= 501 <= nums[i] <= 50The Dynamic Programming approach helps efficiently explore all possible ways to partition the array into subarrays and track the minimal cost achievable. The cost of any subarray is the value of its first element, and in this problem, you can consider breaking down the array such that each partition minimizes the starting costs. You essentially create a function that recursively finds the best division points that lead to the minimal overall start cost of subarrays.
The solution iterates through possible divisions of the array into three subarrays. We continuously check the cost of potential first elements of the subarrays and maintain the minimal cost found. This nested loop approach explores all configurations efficiently given constraints.
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Time complexity: O(n^2), since we're using a nested loop to explore potential divisions.
Space complexity: O(1), as we utilize a constant amount of space outside the input array.
The Greedy approach involves picking partitions of the array based on specific criteria, optimizing each choice to attempt to achieve a globally minimal result. Here, we can focus on selecting ends that guarantee maximizing the cost reduction across splits. Breaking the array in greedy increments may be explored due to small size constraints.
This solution uses a simplified greedy technique, returning the sum of first possible fixed elements of subarrays, assuming that such setup tends to produce generally minimal starting sum. Note: This basic attempt showcases logic's potential, ensuring complete solution iteration might still hold most benefits.
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Time complexity: O(1), owing to a straightforward evaluation.
Space complexity: O(1), fixed due to the lack of dependency on dynamically allocated space.
| Approach | Complexity |
|---|---|
| Dynamic Programming | Time complexity: O(n^2), since we're using a nested loop to explore potential divisions. |
| Greedy Technique | Time complexity: O(1), owing to a straightforward evaluation. |
3013. Divide an Array Into Subarrays With Minimum Cost II | Sliding Window | Multisets • Aryan Mittal • 2,794 views views
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