LeetCode #3424 — MEDIUM

Minimum Cost to Make Arrays Identical

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[-7,9,5]
[7,-2,-5]
2

Example 2

[2,1]
[2,1]
0

Step 02

Core Insight

What unlocks the optimal approach

What does Operation 1 (rearranging subarrays) actually accomplish?
Calculate <code>sum(abs(arr[i] - brr[i]))</code> if you do not use Operation 1.
Calculate <code>sum(abs(arr[i] - brr[i]))</code> after sorting both arrays if you use Operation 1.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3424: Minimum Cost to Make Arrays Identical
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;
    }
}

// Accepted solution for LeetCode #3424: Minimum Cost to Make Arrays Identical
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
}

# Accepted solution for LeetCode #3424: Minimum Cost to Make Arrays Identical
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)

// Accepted solution for LeetCode #3424: Minimum Cost to Make Arrays Identical
impl Solution {
    pub fn min_cost(mut arr: Vec<i32>, mut brr: Vec<i32>, k: i64) -> i64 {
        let c1: i64 = arr
            .iter()
            .zip(&brr)
            .map(|(a, b)| (*a - *b).abs() as i64)
            .sum();

        arr.sort_unstable();
        brr.sort_unstable();

        let c2: i64 = k + arr
            .iter()
            .zip(&brr)
            .map(|(a, b)| (*a - *b).abs() as i64)
            .sum::<i64>();

        c1.min(c2)
    }
}

// Accepted solution for LeetCode #3424: Minimum Cost to Make Arrays Identical
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.