#3424

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Cost to Make Arrays Identical

Minimize cost to make arrays identical by performing operations with given constraints and a greedy strategy.

Problem Statement

You are given two integer arrays, arr and brr, both of length n, and an integer k. The task is to transform arr into brr with the minimum total cost. The allowed operation involves selecting a subarray from arr and rearranging it, costing a fixed amount for each change.

Your objective is to determine the minimal cost to make arr equal to brr. The solution hinges on analyzing how the rearrangement operation impacts the final transformation cost, emphasizing sorting and optimal subarray choices.

Examples

Example 1

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

Output: 13

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

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

Solution Approach

Greedy Approach with Sorting

Start by sorting both arrays arr and brr. This ensures that we handle the smallest values first, optimizing the cost of making them equal. Sorting makes the cost computation more efficient as we can directly match corresponding elements from both arrays.

Cost Calculation Using Invariants

After sorting, calculate the cost to match elements in arr and brr by focusing on the differences between them. The invariant here is that each change between arr[i] and brr[i] is constant for any swap, allowing efficient computation of the minimal cost.

Rearranging Subarrays for Minimum Cost

The transformation is done by rearranging subarrays of arr. The number of such operations is minimized by focusing on making the least expensive changes first. Carefully choosing the smallest swaps ensures that the total cost stays as low as possible.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sorting step, making it O(n log n). The space complexity is O(n) for storing the arrays. These complexities are optimal for solving this problem given the constraints on array size and the cost of operations.

What Interviewers Usually Probe

Can the candidate recognize the greedy approach for minimizing costs?
How efficiently do they handle the sorting step for large inputs?
Do they understand how to minimize operations while ensuring correctness?

Common Pitfalls or Variants

Common pitfalls

Failing to recognize the importance of sorting both arrays first.
Incorrectly handling the cost calculation, especially when dealing with large differences.
Overcomplicating the solution with unnecessary operations, leading to higher costs.

Follow-up variants

What if the arrays are already identical, how does the solution scale?
How would the solution change if k were very large, pushing the problem towards near-linear time complexity?
Can the approach handle edge cases with negative numbers or maximum array lengths efficiently?

FAQ

How does the greedy approach work for the "Minimum Cost to Make Arrays Identical"?

The greedy approach minimizes cost by focusing on sorting both arrays and calculating the differences step-by-step to ensure the cheapest transformations.

What is the key to solving this problem efficiently?

The key is sorting both arrays first, followed by applying a cost calculation based on the differences, ensuring the most efficient solution.

How does the problem handle rearranging subarrays?

Rearranging subarrays minimizes the cost by focusing on matching smaller values first, ensuring the least expensive changes are made first.

What is the time complexity of the solution?

The time complexity is O(n log n) due to the sorting step, which is the most time-consuming part of the solution.

Can negative numbers be part of the arrays in this problem?

Yes, negative numbers are allowed in both arrays, and the solution works efficiently with them by sorting and calculating the differences as usual.

terminal

Solution

Solution 1: Greedy + Sorting

If splitting the array is not allowed, we can directly calculate the sum of absolute differences between the two arrays as the total cost $c_1$. If splitting is allowed, we can divide the array $\textit{arr}$ into $n$ subarrays of length 1, then rearrange them in any order, and compare with array $\textit{brr}$, calculating the sum of absolute differences as the total cost $c_2$. To minimize $c_2$, we can sort both arrays and then calculate the sum of absolute differences. The final result is $\min(c_1, c_2 + k)$.

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class Solution:
    def minCost(self, arr: List[int], brr: List[int], k: int) -> int:
        c1 = sum(abs(a - b) for a, b in zip(arr, brr))
        arr.sort()
        brr.sort()
        c2 = k + sum(abs(a - b) for a, b in zip(arr, brr))
        return min(c1, c2)

Continue Practicing

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