, we can rearrange contiguous subarrays of arr at a fixed cost k, and then adjust each element to match brr using add/subtract operations. The optimal way is to sort both arrays and match elements directly, as rearrangement allows any order.

Approach

Sort both arr and brr.
The cost to rearrange is k (if needed).
For each index, calculate the cost to change arr[i] to brr[i] (absolute difference).
The total cost is the sum of all element changes plus k (if rearrangement is used).
If arr is already equal to brr, no rearrangement cost is needed.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int minimumCost(vector<int>& arr, vector<int>& brr, int k) {\n        vector<int> a = arr, b = brr;\n        sort(a.begin(), a.end());\n        sort(b.begin(), b.end());\n        int ans = k;\n        for (int i = 0; i < a.size(); ++i) ans += abs(a[i] - b[i]);\n        return ans;\n    }\n};","lang":"cpp","html":"<pre class=\"shiki shiki-themes github-light github-dark\" style=\"background-color:#fff;--shiki-dark-bg:#24292e;color:#24292e;--shiki-dark:#e1e4e8\" tabindex=\"0\"><code><span class=

Related Tags