3424. Minimum Cost to Make Arrays Identical

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Description

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

Solutions

Solution 1

Python3

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)

Java

class Solution {
    public long minCost(int[] arr, int[] brr, long k) {
        long c1 = calc(arr, brr);
        Arrays.sort(arr);
        Arrays.sort(brr);
        long c2 = calc(arr, brr) + k;
        return Math.min(c1, c2);
    }

    private long calc(int[] arr, int[] brr) {
        long ans = 0;
        for (int i = 0; i < arr.length; ++i) {
            ans += Math.abs(arr[i] - brr[i]);
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long minCost(vector<int>& arr, vector<int>& brr, long long k) {
        auto calc = [&](vector<int>& arr, vector<int>& brr) {
            long long ans = 0;
            for (int i = 0; i < arr.size(); ++i) {
                ans += abs(arr[i] - brr[i]);
            }
            return ans;
        };
        long long c1 = calc(arr, brr);
        ranges::sort(arr);
        ranges::sort(brr);
        long long c2 = calc(arr, brr) + k;
        return min(c1, c2);
    }
};

Go

func minCost(arr []int, brr []int, k int64) int64 {
	calc := func(a, b []int) (ans int64) {
		for i := range a {
			ans += int64(abs(a[i] - b[i]))
		}
		return
	}
	c1 := calc(arr, brr)
	sort.Ints(arr)
	sort.Ints(brr)
	c2 := calc(arr, brr) + k
	return min(c1, c2)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

function minCost(arr: number[], brr: number[], k: number): number {
    const calc = (a: number[], b: number[]) => {
        let ans = 0;
        for (let i = 0; i < a.length; ++i) {
            ans += Math.abs(a[i] - b[i]);
        }
        return ans;
    };
    const c1 = calc(arr, brr);
    arr.sort((a, b) => a - b);
    brr.sort((a, b) => a - b);
    const c2 = calc(arr, brr) + k;
    return Math.min(c1, c2);
}