<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 countFruits(self, basket):
        counts = defaultdict(int)
        for fruit in basket:
            counts[fruit] += 1
        return counts

    def minCost(self, basket1: List[int], basket2: List[int]) -> int:
        abs_min = min(min(basket1), min(basket2))
        counts1 = self.countFruits(basket1)
        counts2 = self.countFruits(basket2)
        all_fruits = counts1.keys() | counts2.keys()
        heap = []
        for fruit in all_fruits:
            diff = abs(counts1[fruit] - counts2[fruit])
            if diff % 2 == 1:
                # baskets are of equal size, so can't have an odd number
                # of any fruit
                return -1
            for _ in range(diff//2):
                # Half of these fruits must live in each array, so 
                # we can use half of them to swap items with the other array
                heappush(heap, fruit)
        
        cost = 0
        for _ in range(len(heap)//2):
            cost += min(2 * abs_min, heappop(heap))
        return cost


        
639