Rearranging Fruits
<p>You have two fruit baskets containing <code>n</code> fruits each. You are given two <strong>0-indexed</strong> integer arrays <code>basket1</code> and <code>basket2</code> representing the cost of fruit in each basket. You want to make both baskets <strong>equal</strong>. To do so, you can use the following operation as many times as you want:</p>

<ul>
	<li>Choose two indices <code>i</code> and <code>j</code>, and swap the <code>i<sup><font size="1">th</font></sup></code> fruit of <code>basket1</code> with the <code>j<sup><font size="1">th</font></sup></code> fruit of <code>basket2</code>.</li>
	<li>The cost of the swap is <code>min(basket1[i], basket2[j])</code>.</li>
</ul>

<p>Two baskets are considered equal if sorting them according to the fruit cost makes them exactly the same baskets.</p>

<p>Return <em>the minimum cost to make both the baskets equal or </em><code>-1</code><em> if impossible.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> basket1 = [4,2,2,2], basket2 = [1,4,1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Swap index 1 of basket1 with index 0 of basket2, which has cost 1. Now basket1 = [4,1,2,2] and basket2 = [2,4,1,2]. Rearranging both the arrays makes them equal.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> basket1 = [2,3,4,1], basket2 = [3,2,5,1]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be shown that it is impossible to make both the baskets equal.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>basket1.length == basket2.length</code></li>
	<li><code>1 &lt;= basket1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= basket1[i], basket2[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Greedy; Sort; Array; Hash Table
class Solution {
public:
    long long minCost(vector<int>& basket1, vector<int>& basket2) {
        int n = basket1.size();
        unordered_map<int, int> cnt;
        for (int i = 0; i < n; ++i) {
            cnt[basket1[i]]++;
            cnt[basket2[i]]--;
        }
        int mi = 1 << 30;
        vector<int> nums;
        for (auto& [x, v] : cnt) {
            if (v % 2) {
                return -1;
            }
            for (int i = abs(v) / 2; i; --i) {
                nums.push_back(x);
            }
            mi = min(mi, x);
        }
        ranges::sort(nums);
        int m = nums.size();
        long long ans = 0;
        for (int i = 0; i < m / 2; ++i) {
            ans += min(nums[i], mi * 2);
        }
        return ans;
    }
};