<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>

from sortedcontainers import SortedList

class Solution:
    def minCost(self, basket1: List[int], basket2: List[int]) -> int:

        x, y = SortedList(), SortedList()
        a, b = Counter(basket1), Counter(basket2)
        for k in a.keys() | b.keys():
            i, j = a[k], b[k]
            if (i + j) % 2:
                return -1
            if i < j: 
                n = (i + j) // 2
                for _ in range(j - n):
                    y.add(k)
            if i > j: 
                n = (i + j) // 2
                for _ in range(i - n):
                    x.add(k)

        mn = min(a.keys() | b.keys())
        p = sum(n == mn for n in x + y)
        ans = k = mn * ((len(x) * 2) - p)
        for _ in range(len(x)):
            if x[-1] > y[-1]:
                k += y[0] - mn * (2 - (y[0] == mn))
                y.pop(0)
                x.pop(-1)
            else:
                k += x[0] - mn * (2 - (x[0] == mn))
                x.pop(0)
                y.pop(-1)
            ans = min(ans, k)

        return ans
44800