<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>Chose two indices <code>i</code> and <code>j</code>, and swap the <code>i<font size="1">th</font>&nbsp;</code>fruit of <code>basket1</code> with the <code>j<font size="1">th</font></code>&nbsp;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]&nbsp;&lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minCost(self, basket1: List[int], basket2: List[int]) -> int:
        # It is possible only when the frequency of each fruit cost is even.
        # We can first count the total frequeny of fruit cost, and derive the expected fruit cost frequency in each basket.
        # We can sort out the redundant fruit cost in each basket. Note that a given fruit cost cannot be redundant in both baskets.
        # For the smallest redundant element, we have two options:
        # 1. Swapping it with the largest redundant element with the largest one in other side. The cost is the smallest redundant elemnt.
        # 2. Swapping it with the smallest balanced element in other side first, and use this element to swap the largest one in other side. The cost is 2 * smallest balanced element.
        counter1 = collections.Counter(basket1)
        counter2 = collections.Counter(basket2)
        total_counter = counter1 + counter2
        if any(total_counter[k] % 2 == 1 for k in total_counter):
            return -1
        del_counter = collections.Counter()
        delta = 0
        min_cost = float("inf")
        for k in total_counter:
            min_cost = min(min_cost, k)
            e = total_counter[k] // 2
            if counter1[k] > e:
                del_counter[k] = counter1[k] - e
                delta += counter1[k] - e
            elif counter2[k] > e:
                del_counter[k] = counter2[k] - e
                delta += counter2[k] - e
        if delta == 0:
            # This is already well arranged.
            return 0
        ans = 0
        cur = 0
        target = delta >> 1
        for k in sorted(del_counter):
            if cur + del_counter[k] >= target:
                return ans + min(k, 2 * min_cost) * (target - cur)
            cur += del_counter[k]
            ans += del_counter[k] * min(k, 2 * min_cost)
        # This is unreachable
        return -1
        
611