LeetCode #2499 — HARD

Minimum Total Cost to Make Arrays Unequal

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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
Patterns Used
🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,2,3,4,5]
[1,2,3,4,5]

Example 2

[2,2,2,1,3]
[1,2,2,3,3]

Example 3

[1,2,2]
[1,2,2]
Step 02

Core Insight

What unlocks the optimal approach

  • How can we check which indices of <code>nums1</code> will be considered for swapping? How to minimize the number of such operations?
  • It can be seen that greedily swapping values of indices where <code>nums1[i] == nums2[i]</code> is the most optimal choice. How many values cannot be swapped this way?
  • Find which indices we will swap these remaining values with, and if there are enough such indices.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #2499: Minimum Total Cost to Make Arrays Unequal
class Solution {
    public long minimumTotalCost(int[] nums1, int[] nums2) {
        long ans = 0;
        int same = 0;
        int n = nums1.length;
        int[] cnt = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            if (nums1[i] == nums2[i]) {
                ans += i;
                ++same;
                ++cnt[nums1[i]];
            }
        }
        int m = 0, lead = 0;
        for (int i = 0; i < cnt.length; ++i) {
            int t = cnt[i] * 2 - same;
            if (t > 0) {
                m = t;
                lead = i;
                break;
            }
        }
        for (int i = 0; i < n; ++i) {
            if (m > 0 && nums1[i] != nums2[i] && nums1[i] != lead && nums2[i] != lead) {
                ans += i;
                --m;
            }
        }
        return m > 0 ? -1 : ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n log n)
Space
O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.

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