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Minimum Cost to Split an Array #4 Most Asked

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ArraysDynamic Programming

You are given an integer array nums and an integer k.

Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split.

Let trimmed(subarray) be the version of the subarray where all numbers which appear only once are removed.

  • For example, trimmed([3,1,2,4,3,4]) = [3,4,3,4].

The importance value of a subarray is k + trimmed(subarray).length.

  • For example, if a subarray is [1,2,3,3,3,4,4], then trimmed([1,2,3,3,3,4,4]) = [3,3,3,4,4].The importance value of this subarray will be k + 5.

Return the minimum possible cost of a split of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,1,2,1,3,3], k = 2
Output: 8
Explanation: We split nums to have two subarrays: [1,2], [1,2,1,3,3].
The importance value of [1,2] is 2 + (0) = 2.
The importance value of [1,2,1,3,3] is 2 + (2 + 2) = 6.
The cost of the split is 2 + 6 = 8. It can be shown that this is the minimum possible cost among all the possible splits.

Example 2:

Input: nums = [1,2,1,2,1], k = 2
Output: 6
Explanation: We split nums to have two subarrays: [1,2], [1,2,1].
The importance value of [1,2] is 2 + (0) = 2.
The importance value of [1,2,1] is 2 + (2) = 4.
The cost of the split is 2 + 4 = 6. It can be shown that this is the minimum possible cost among all the possible splits.

Example 3:

Input: nums = [1,2,1,2,1], k = 5
Output: 10
Explanation: We split nums to have one subarray: [1,2,1,2,1].
The importance value of [1,2,1,2,1] is 5 + (3 + 2) = 10.
The cost of the split is 10. It can be shown that this is the minimum possible cost among all the possible splits.

Constraints:

  • 1 <= nums.length <= 1000
  • 0 <= nums[i] < nums.length
  • 1 <= k <= 109

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the size of the `nums` array and the value of `k`? Specifically, how large can `n` (the length of `nums`) be, and can `k` be negative or zero?
  2. Could you clarify the definition of the cost of a subarray? For a subarray like `[1, 1, 2, 1]`, is the 'length of its trimmed version' the number of unique elements with duplicates (which is 1 for the element `1`), or the total count of non-unique elements (which would be 2 extra `1`s)?
  3. What is the expected range of values for the integers within the `nums` array? Can they be positive, negative, or zero, and what is their maximum possible value?
  4. What should be the return value if the input array `nums` is empty? Should it be 0, or is this an invalid case?
  5. If the array is split into `m` subarrays, is the split cost `k * (m - 1)` added to the total cost? For example, if there's no split (the array is a single subarray, so m=1), is the added cost `k * 0 = 0`?

Brute Force Solution

Approach

To find the cheapest way to split a sequence of numbers, the brute force approach is to consider every single possible way to cut it up. We then calculate the total cost for each complete split and simply pick the one with the lowest cost.

Here's how the algorithm would work step-by-step:

  1. Imagine the sequence of numbers laid out in a line.
  2. First, consider making the very first cut after the first number. Now you have two pieces: the first number by itself, and the rest of the sequence.
  3. For the remaining part of the sequence, repeat the process: try cutting it after its first number, then after its second, and so on, exploring every possibility recursively.
  4. After exploring all the cuts starting with the first number, go back to the original sequence and try making the first cut after the second number instead.
  5. Again, for the remaining part of the sequence, explore all the ways it can be cut up.
  6. Continue this process, trying every possible location for the first cut, all the way to the end of the sequence.
  7. As you create each complete split of the original sequence into smaller parts, calculate its total cost by adding up the costs of all the individual parts.
  8. Keep track of the minimum cost you've seen so far across all the complete splits you've examined.
  9. After checking every single possible way to split the sequence, the lowest cost you found is your answer.

Code Implementation

def min_cost_split_brute_force(numbers, k_value):

    def get_trimmed_length(subarray):
        counts = {}
        for number in subarray:
            counts[number] = counts.get(number, 0) + 1
        
        trimmed_len = 0
        for number in subarray:
            if counts[number] > 1:
                trimmed_len += 1
        return trimmed_len

    def get_subarray_cost(subarray):
        return k_value + get_trimmed_length(subarray)

    memoization_cache = {}

    def find_minimum_cost_from_index(start_index):
        if start_index >= len(numbers):
            return 0

        if start_index in memoization_cache:
            return memoization_cache[start_index]

        # Initialize minimum cost for the split starting at `start_index` to a very large number.

        minimum_split_cost = float('inf')

        # Iterate through all possible end points to define the first subarray of this recursive call.

        for end_index in range(start_index, len(numbers)):
            current_subarray = numbers[start_index : end_index + 1]
            cost_of_this_subarray = get_subarray_cost(current_subarray)

            # Recursively find the minimum cost for the rest of the array after the current split.

            cost_of_rest_of_array = find_minimum_cost_from_index(end_index + 1)

            total_cost_for_this_path = cost_of_this_subarray + cost_of_rest_of_array
            minimum_split_cost = min(minimum_split_cost, total_cost_for_this_path)
        
        # Store the result to avoid recomputing the minimum cost for this starting index.

        memoization_cache[start_index] = minimum_split_cost
        return minimum_split_cost

    return find_minimum_cost_from_index(0)

Big(O) Analysis

Time Complexity
O(2^n)The brute force approach explores every single possible way to partition the array. For an array of size n, at each of the n-1 positions between elements, we have two choices: either make a cut or not make a cut. This leads to 2^(n-1) possible ways to split the array. For each split, we have to iterate through the parts to calculate the cost, which takes O(n) time. The total number of operations is therefore proportional to n * 2^(n-1), which simplifies to an exponential time complexity of O(2^n).
Space Complexity
O(N)The brute-force approach described uses recursion to explore all possible splits. Each recursive call represents a decision point for a smaller subproblem, starting from a specific index. Since the algorithm can make a cut after the first number, then the second, and so on, the recursion can go as deep as the length of the input sequence, N. This sequence of nested function calls creates a call stack, and the maximum depth of this stack will be proportional to N, resulting in O(N) auxiliary space.

Optimal Solution

Approach

To find the cheapest way to split the list, we can solve it piece by piece. We figure out the best cost for splitting smaller initial portions of the list and use those answers to efficiently calculate the best cost for larger portions, avoiding repetitive work.

Here's how the algorithm would work step-by-step:

  1. Imagine you need to decide where the very last split should be. The final sub-list could be the last number by itself, the last two numbers together, the last three, and so on.
  2. For each of these possibilities for the final sub-list, you need to calculate its individual cost.
  3. The total cost for a particular choice of the final sub-list is its own cost plus the best possible cost for splitting all the numbers that came before it.
  4. Since we need the pre-calculated best cost for the part that comes before, it makes sense to start from the beginning of the list and work our way forward.
  5. First, find the best cost for splitting just the first number, then the first two, then the first three, and so on.
  6. To find the best cost for a portion of the list (say, the first ten numbers), you try every possible final sub-list: just the tenth number, the ninth and tenth together, etc.
  7. For each of these potential final sub-lists, you add its cost to the already-known best cost of the part preceding it. For example, if you're testing the ninth and tenth numbers as the final group, you'd add their cost to the best cost you've already found for the first eight numbers.
  8. You then pick the option that gives you the lowest total cost. This becomes the new 'best cost' for splitting the first ten numbers.
  9. By repeating this process until you reach the end of the full list, you guarantee you've found the overall minimum cost without having to check every single possible split from scratch.

Code Implementation

from collections import Counter

def min_cost(nums, k):
    array_length = len(nums)
    # This DP array stores the minimum cost to split the subarray from the start up to index i-1.
    min_cost_dp = [float('inf')] * (array_length + 1)
    min_cost_dp[0] = 0

    # Iterate through each possible end point of a split, building up the solution.
    for end_index in range(1, array_length + 1):
        current_min_total_cost = float('inf')
        frequency_counter = Counter()
        trimmed_length = 0

        # Consider all possible start points for the final subarray ending at end_index-1.
        for start_index in range(end_index - 1, -1, -1):
            current_num = nums[start_index]
            frequency_counter[current_num] += 1

            # Calculate the trimmed_length for the current subarray [start_index...end_index-1].
            if frequency_counter[current_num] == 2:
                trimmed_length += 2
            elif frequency_counter[current_num] > 2:
                trimmed_length += 1

            cost_of_last_subarray = k + trimmed_length
            # Combine the cost of the final subarray with the optimal cost of the preceding part.
            total_cost_for_this_split = min_cost_dp[start_index] + cost_of_last_subarray
            current_min_total_cost = min(current_min_total_cost, total_cost_for_this_split)
        
        min_cost_dp[end_index] = current_min_total_cost

    return min_cost_dp[array_length]

Big(O) Analysis

Time Complexity
O(n²)The solution calculates the minimum cost for splitting the array prefix of every possible length from 1 to n. To find the cost for a prefix of length i, it iterates through all possible split points j before i, creating a final subarray from j to i. This results in a nested loop structure where the outer loop runs n times (for each prefix length i) and the inner loop runs on average n/2 times (for each possible last split point j). The total number of operations is proportional to n * n, which simplifies to O(n²).
Space Complexity
O(N)The algorithm calculates the best cost for splitting progressively larger portions of the list, from the first number up to the first N numbers. The plain English explanation states, "you'd add their cost to the already-known best cost of the part preceding it," which implies that we need to store these intermediate best costs. This requires an auxiliary data structure, typically a DP array, of size N+1 to save the minimum cost for splitting the first i elements. Therefore, the extra space used grows linearly with the size of the input list, N.

Edge Cases

Empty input array `nums`
How to Handle:
The cost should be 0 as there are no subarrays to form and no splits to make.
Single element array `nums`
How to Handle:
The only split is the array itself, so the cost is its subarray cost (1) with zero splits.
The value of `k` is zero
How to Handle:
The cost of splitting becomes free, so the optimal solution may involve many small subarrays.
The value of `k` is very large
How to Handle:
A large `k` heavily penalizes splits, making the optimal solution likely to be splitting the array into just one subarray.
All elements in `nums` are unique
How to Handle:
The trimmed length of any subarray is always its actual length, simplifying the subarray cost calculation to just `2 * length`.
All elements in `nums` are identical
How to Handle:
The trimmed length of any subarray of length `L > 1` is `L - (L-1) = 1`, making its cost `L + 1`.
Maximum input size for `nums` (e.g., N=1000)
How to Handle:
The solution must be more efficient than exponential, likely an O(N^2) dynamic programming approach to avoid a timeout.
Input `nums` contains negative numbers or zeros
How to Handle:
The values of the numbers do not affect the logic, only their frequencies, so the algorithm handles these cases correctly.
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