Minimum Cost to Make Arrays Identical

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ArraysGreedy Algorithms

You are given two integer arrays arr and brr of length n, and an integer k. You can perform the following operations on arr any number of times:

Split arr into any number of contiguous subarrays and rearrange these subarrays in any order. This operation has a fixed cost of k.
Choose any element in arr and add or subtract a positive integer x to it. The cost of this operation is x.

Return the minimum total cost to make arr equal to brr.

Example 1:

Input: arr = [-7,9,5], brr = [7,-2,-5], k = 2

Output: 13

Explanation:

Split arr into two contiguous subarrays: [-7] and [9, 5] and rearrange them as [9, 5, -7], with a cost of 2.
Subtract 2 from element arr[0]. The array becomes [7, 5, -7]. The cost of this operation is 2.
Subtract 7 from element arr[1]. The array becomes [7, -2, -7]. The cost of this operation is 7.
Add 2 to element arr[2]. The array becomes [7, -2, -5]. The cost of this operation is 2.

The total cost to make the arrays equal is 2 + 2 + 7 + 2 = 13.

Example 2:

Input: arr = [2,1], brr = [2,1], k = 0

Output: 0

Explanation:

Since the arrays are already equal, no operations are needed, and the total cost is 0.

Constraints:

1 <= arr.length == brr.length <= 10⁵
0 <= k <= 2 * 10¹⁰
-10⁵ <= arr[i] <= 10⁵
-10⁵ <= brr[i] <= 10⁵

Try coding it in LeetCode

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:

What is the range of values within the arrays? Can they be negative, zero, or non-integer?
What constitutes a 'cost' and how is it calculated based on the operation performed on array elements to make them identical? Can you give an example?
If it's impossible to make the arrays identical, what value should I return?
Are the arrays guaranteed to be of the same length?
Are there any constraints on the types of operations I can perform on the array elements to make them identical?

Brute Force Solution

Approach

The basic idea is to explore every possible way to make the collections identical. We will try every possible change to each collection, check if the collections are now identical, and then keep track of the cheapest set of changes we found to make them identical.

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

Start with the first collection and consider all the different modifications you can make to its individual elements.
For each of those modifications, consider all the different modifications you can make to the second collection's individual elements.
For each possible combination of changes, check if the collections are identical now.
If they are identical, calculate the cost of the modifications we made.
Keep track of the minimum cost we've seen so far.
Repeat this process, considering every possible combination of modifications to both collections until we have tried everything.
The lowest cost we found after trying all possibilities is the answer.

Code Implementation

def min_cost_to_make_arrays_identical_brute_force(first_array, second_array, modification_cost):
    minimum_cost = float('inf')

    # Iterate through all possible modifications of the first array
    for i in range(3 ** len(first_array)):
        temp_first_array = first_array[:]
        current_cost = 0
        temp_index = i

        for index in range(len(first_array)):
            modification = temp_index % 3
            temp_index //= 3

            if modification == 1:
                temp_first_array[index] += 1
                current_cost += modification_cost
            elif modification == 2:
                temp_first_array[index] -= 1
                current_cost += modification_cost

        # Iterate through all possible modifications of the second array
        for j in range(3 ** len(second_array)):
            temp_second_array = second_array[:]
            inner_cost = 0
            inner_index = j

            for index in range(len(second_array)):
                modification = inner_index % 3
                inner_index //= 3

                if modification == 1:
                    temp_second_array[index] += 1
                    inner_cost += modification_cost
                elif modification == 2:
                    temp_second_array[index] -= 1
                    inner_cost += modification_cost

            # Check if the arrays are identical after modifications
            if temp_first_array == temp_second_array:

                # Update minimum cost if the current cost is lower
                minimum_cost = min(minimum_cost, current_cost + inner_cost)

    if minimum_cost == float('inf'):
        return -1
    else:
        return minimum_cost

Big(O) Analysis

Time Complexity

O(k^(2n)) – Let n be the size of each collection and k be the number of possible modifications to each element. For each of the n elements in the first collection, we explore k possible modifications. Similarly, for each of the n elements in the second collection, we explore k possible modifications. Thus, we have k^n possibilities for each collection. Since we consider all combinations of modifications to both collections, the total number of combinations is k^n * k^n = k^(2n). Therefore, the time complexity is O(k^(2n)).

Space Complexity

O(1) – The provided solution explores all possible modifications to the input arrays to find the minimum cost. Critically, it only keeps track of the minimum cost seen so far and doesn't store intermediate states or large collections of modifications. Therefore, the space complexity is dominated by the storage of a few variables like the current minimum cost and possibly loop counters, which takes constant space regardless of the input size (number of elements in the arrays). No auxiliary data structures that scale with the input size are used.

Optimal Solution

Approach

The most efficient way to solve this is to focus on modifying each list to match a single, central value. We find that central value strategically, and then calculate the cost of changing each list to match it.

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

First, find the value that appears most often when both lists are combined. If there are ties, pick any of the most frequent values.
Calculate the cost to change all numbers in the first list to match this most frequent value.
Calculate the cost to change all numbers in the second list to match this most frequent value.
Add these two costs together. This is the minimum cost to make both lists identical.
The key insight is that changing both lists to this combined most frequent number minimizes changes, and therefore the total cost.

Code Implementation

def min_cost_to_make_arrays_identical(first_list, second_list):
    combined_list = first_list + second_list
    
    # Find the most frequent number
    most_frequent_number = max(set(combined_list), key=combined_list.count)

    cost_first_list = 0
    for number in first_list:
        if number != most_frequent_number:
            cost_first_list += 1

    cost_second_list = 0
    for number in second_list:
        if number != most_frequent_number:

            cost_second_list += 1

    # The total cost is the sum of the individual costs
    total_cost = cost_first_list + cost_second_list
    return total_cost

Big(O) Analysis

Time Complexity

O(n) – Finding the most frequent element requires iterating through both lists once each, where n is the total number of elements in both lists combined. Calculating the cost to change each list to this most frequent value also requires iterating through each list once. Therefore, the algorithm performs a constant number of linear scans of the combined lists, resulting in O(n) time complexity.

Space Complexity

O(N) – The algorithm's space complexity is dominated by the need to determine the most frequent value when both lists are combined. This requires creating a frequency map, which, in the worst case, could store a count for each unique element in both lists. Therefore, the frequency map could potentially store up to N distinct elements, where N is the total number of elements in both lists. Hence, the auxiliary space required grows linearly with the input size N.

Edge Cases

Case	How to Handle
Empty arrays A or B	Return 0 since no operations are needed if either array is empty.
Arrays A and B have different lengths	Return -1, indicating that the arrays cannot be made identical via the defined operations.
Arrays A and B are already identical	Return 0 since no cost is incurred.
Arrays A and B contain negative numbers or zeros	The cost function should handle negative numbers correctly by adding the absolute differences.
Large array sizes potentially leading to integer overflow in cost calculations	Use long long integer type to store the cost during calculations.
The cost of changing an element can be zero.	The minimum cost between array elements will be selected despite a zero-cost change being possible
No sequence of operations will ever cause the arrays to be identical	The algorithm needs a mechanism to detect this impossibility and return an appropriate error, such as -1, perhaps when the sum of elements is fundamentally different.
Integer overflow when calculating the absolute difference between elements	Explicitly cast the elements to a larger integer type like 'long long' before calculating the difference.

Empty arrays A or B

How to Handle:

Return 0 since no operations are needed if either array is empty.

Arrays A and B have different lengths

How to Handle:

Return -1, indicating that the arrays cannot be made identical via the defined operations.

Arrays A and B are already identical

How to Handle:

Return 0 since no cost is incurred.

Arrays A and B contain negative numbers or zeros

How to Handle:

The cost function should handle negative numbers correctly by adding the absolute differences.

Large array sizes potentially leading to integer overflow in cost calculations

How to Handle:

Use long long integer type to store the cost during calculations.

The cost of changing an element can be zero.

How to Handle:

The minimum cost between array elements will be selected despite a zero-cost change being possible

No sequence of operations will ever cause the arrays to be identical

How to Handle:

The algorithm needs a mechanism to detect this impossibility and return an appropriate error, such as -1, perhaps when the sum of elements is fundamentally different.

Integer overflow when calculating the absolute difference between elements

How to Handle:

Explicitly cast the elements to a larger integer type like 'long long' before calculating the difference.