<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:
        c1 = defaultdict(int)
        c2 = defaultdict(int)
        count = defaultdict(int)
        unq = set()

        for i in basket1:
            c1[i] += 1
            count[i] += 1
            unq.add(i)
        for i in basket2:
            c2[i] += 1
            count[i] += 1
            unq.add(i)    
        for i in count.values():
            if i % 2 != 0:
                return -1
        
        diff1 = []
        diff2 = []
        unq = sorted(list(unq))
        # print(unq)
        min_elt = unq[0]
        for i in unq:
            f1 = c1[i]
            f2 = c2[i]

            if f1 == f2:
                continue
            
            diff = abs(f1-f2)

            if f1>f2:
                for j in range(0, diff//2):
                    diff1.append(i)

            else:
                for j in range(0, diff//2):
                    diff2.append(i)
        
        if len(diff1) != len(diff2):
            return -1
        
        diff2 = diff2[::-1]
        cost = 0
        # print(diff1, diff2)
        for i in range(len(diff1)):
            cost += min(min_elt*2, min(diff1[i], diff2[i]))
        return cost
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