Minimum Total Cost to Make Arrays Unequal

Hard

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ArraysGreedy Algorithms

You are given two 0-indexed integer arrays nums1 and nums2, of equal length n.

In one operation, you can swap the values of any two indices of nums1. The cost of this operation is the sum of the indices.

Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n - 1 after performing all the operations.

Return the minimum total cost such that nums1 and nums2 satisfy the above condition. In case it is not possible, return -1.

Example 1:

Input: nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5]
Output: 10
Explanation: 
One of the ways we can perform the operations is:
- Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5]
- Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5].
- Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4].
We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10.
Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10.

Example 2:

Input: nums1 = [2,2,2,1,3], nums2 = [1,2,2,3,3]
Output: 10
Explanation: 
One of the ways we can perform the operations is:
- Swap values at indices 2 and 3, incurring cost = 2 + 3 = 5. Now, nums1 = [2,2,1,2,3].
- Swap values at indices 1 and 4, incurring cost = 1 + 4 = 5. Now, nums1 = [2,3,1,2,2].
The total cost needed here is 10, which is the minimum possible.

Example 3:

Input: nums1 = [1,2,2], nums2 = [1,2,2]
Output: -1
Explanation: 
It can be shown that it is not possible to satisfy the given conditions irrespective of the number of operations we perform.
Hence, we return -1.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
1 <= nums1[i], nums2[i] <= n

Try coding it in LeetCode

Minimum Total Cost to Make Arrays Unequal

Hard

Asked by:

5 views

Topics:

ArraysGreedy Algorithms

You are given two 0-indexed integer arrays nums1 and nums2, of equal length n.

In one operation, you can swap the values of any two indices of nums1. The cost of this operation is the sum of the indices.

Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n - 1 after performing all the operations.

Return the minimum total cost such that nums1 and nums2 satisfy the above condition. In case it is not possible, return -1.

Example 1:

Input: nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5]
Output: 10
Explanation: 
One of the ways we can perform the operations is:
- Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5]
- Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5].
- Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4].
We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10.
Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10.

Example 2:

Input: nums1 = [2,2,2,1,3], nums2 = [1,2,2,3,3]
Output: 10
Explanation: 
One of the ways we can perform the operations is:
- Swap values at indices 2 and 3, incurring cost = 2 + 3 = 5. Now, nums1 = [2,2,1,2,3].
- Swap values at indices 1 and 4, incurring cost = 1 + 4 = 5. Now, nums1 = [2,3,1,2,2].
The total cost needed here is 10, which is the minimum possible.

Example 3:

Input: nums1 = [1,2,2], nums2 = [1,2,2]
Output: -1
Explanation: 
It can be shown that it is not possible to satisfy the given conditions irrespective of the number of operations we perform.
Hence, we return -1.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
1 <= nums1[i], nums2[i] <= n

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 are the constraints on the size of the input arrays nums1 and nums2?
Can the values within nums1 and nums2 be negative, zero, or have other constraints?
If it's impossible to make nums1[i] != nums2[i] for all i, is -1 the only acceptable return value or are there other ways to signal failure?
Are there any guarantees about the relationship between the values in nums1 and nums2? For example, are they guaranteed to be integers?
If multiple solutions exist that achieve the minimum total cost, is any one of them acceptable?

Brute Force Solution

Approach

The brute force approach to making two collections of numbers 'unequal' involves checking every possible way to rearrange one of the collections. We want to find the arrangement that costs the least, where the cost is determined by the pairing between the two collections.

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

Consider all possible arrangements of one of the number collections.
For each arrangement, pair up the numbers from the first collection with the rearranged second collection in the same order.
Compare each pair of numbers; if any pair has the same number, this arrangement is invalid, and we move to the next arrangement.
If all pairs in the current arrangement are different, then calculate the total cost for this arrangement. The cost for a pair is the product of the two numbers.
Remember the smallest total cost you've found so far.
After trying all possible arrangements, the smallest total cost you remembered is the answer.

Code Implementation

import itertools

def minimum_total_cost_to_make_arrays_unequal(first_array, second_array):
    minimum_cost = float('inf')
    for permutation in itertools.permutations(second_array):
        total_cost_for_permutation = 0
        is_valid_permutation = True
        
        # The permutation needs to be checked for validity
        for index in range(len(first_array)):
            if first_array[index] == permutation[index]:
                is_valid_permutation = False
                break
                
        if is_valid_permutation:
            # Calculate total cost only for valid permutations
            for index in range(len(first_array)):
                total_cost_for_permutation += first_array[index] * permutation[index]

            # Update minimum cost if this permutation is cheaper
            minimum_cost = min(minimum_cost, total_cost_for_permutation)

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

Big(O) Analysis

Time Complexity

O(n * n!) – The algorithm iterates through all possible permutations of one of the input arrays of size n. Generating all permutations takes O(n!) time. For each permutation, the algorithm then iterates through the n elements of the array to calculate the cost, which includes pair checking. The pair checking step which involves n operations dominates the cost inside the permutation loop. Therefore, the overall time complexity is O(n * n!).

Space Complexity

O(N) – The brute force approach involves generating all possible arrangements (permutations) of one of the number collections. To generate these permutations, a recursive algorithm or an iterative approach that stores intermediate permutations is generally employed. The depth of the recursion or the size of the intermediate permutations stored will be proportional to N, where N is the number of elements in the collections being permuted. Therefore, the auxiliary space is O(N).

Optimal Solution

Approach

The goal is to make two lists unequal while minimizing the total cost. We'll focus on matching identical numbers in the two lists and then swapping them efficiently, prioritizing the swaps that cost the least.

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

First, count how many times each number appears in both lists.
Identify the numbers that appear in the same spot in both lists. These are the ones causing problems.
Now, look at the numbers that are *not* in the same spot in both lists. These are our potential 'swappers'.
For each problematic number, find the 'swapper' with the lowest associated cost (from either list).
If there are more problematic numbers than available 'swappers', find the lowest possible cost among *all* numbers, and repeatedly use that number as the swapper.
The total cost is the sum of the costs of all the swaps we've made.

Code Implementation

def min_total_cost(numbers1, numbers2, cost):
    number_of_elements = len(numbers1)
    unequal_cost = 0

    # Identify indices where numbers are the same in both lists
    problematic_indices = [i for i in range(number_of_elements) if numbers1[i] == numbers2[i]]

    # Find potential swappers (indices where numbers are different)
    potential_swappers = [i for i in range(number_of_elements) if numbers1[i] != numbers2[i]]

    # Prioritize swaps with existing available swappers.
    for index in problematic_indices[:]:
        if potential_swappers:
            min_swap_cost_index = potential_swappers.pop(0)
            unequal_cost += cost[index] + cost[min_swap_cost_index]
            problematic_indices.remove(index)

    # If more problematic indices remain, use the overall minimum cost.
    if problematic_indices:
        #Determine the minimum cost among all elements.
        min_cost = min(cost)

        remaining_swaps = len(problematic_indices)
        unequal_cost += remaining_swaps * 2 * min_cost

    return unequal_cost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations in this solution involve counting element frequencies (O(n) using a hash map), identifying elements at the same indices (O(n)), and then finding optimal 'swappers' for the problematic numbers. The crucial step is finding the minimum cost 'swapper', which, if implemented efficiently using a priority queue (heap), takes O(log n) for each problematic number. Since we might have up to 'n' problematic numbers, the overall time complexity becomes O(n log n). Specifically, operations such as sorting and priority queue operations can greatly impact the time complexity of the code.

Space Complexity

O(N) – The algorithm counts the occurrences of each number in both lists, implying the use of hash maps or dictionaries to store these counts. The size of these hash maps would be proportional to the number of unique elements in the input lists, which in the worst case could be N (where N is the length of the input lists). Additionally, temporary lists may be used to identify and store the 'problematic numbers' and potential 'swappers', contributing O(N) space. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input arrays nums1 or nums2	Return 0 if both are null/empty, or -1 if only one is null/empty, or if they have different lengths.
Arrays of length 1	If nums1[0] == nums2[0], return -1; otherwise, return 0 (no swap needed).
Arrays of length 2 where a single swap makes them unequal	Check if nums1[0] == nums2[0] and nums1[1] == nums2[1]; if so, the cost is either 0 or 1 and pick the swap index.
Arrays where nums1[i] == nums2[i] for all i	Find a minimum cost swap to make all elements unequal, otherwise return -1 if no possible swap exist.
Arrays where no swap is necessary (nums1[i] != nums2[i] for all i)	Return 0 since no swaps are required.
Arrays with large lengths and integer overflow during cost calculation	Use long data type to store the cost and intermediate sums to prevent potential integer overflow.
Case where the minimum cost involves multiple swap combinations.	Ensure the algorithm explores various swap possibilities to find the absolute minimum cost and not a local minimum.
Input arrays containing duplicate values, and certain values prevent the arrays from being unequal.	Check all possible valid swap operations that yield the minimum cost or report impossibility.