<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, c2 = dict(Counter(basket1)), dict(Counter(basket2))
        
        # freq map from b1 to b2, and from b2 to b1
        m12, m21 = {}, {}
        for k in c1:
            if c1[k] == c2.get(k, 0): continue
            tot = c1[k] + c2.get(k, 0)
            if tot % 2:
                return -1
            if c1[k] > c2.get(k, 0):
                m12[k] = c1[k] - tot // 2
            else:
                m21[k] = c2.get(k, 0) - tot // 2

        for k in c2:
            if k in m12 or k in m21 or c2[k] == c1.get(k, 0): continue
            tot = c2[k] + c1.get(k, 0)
            if tot % 2:
                return -1
            if c2[k] > c1.get(k, 0):
                m21[k] = c2[k] - tot // 2
            else:
                m12[k] = c1.get(k, 0) - tot // 2
        
        # impossible to swap if unmatched swap counts
        if sum(m12.values()) != sum(m21.values()): return -1
        
        # number to swap sorted
        pos, neg = sorted(m12), sorted(m21)
        
        # minimum element for two backets
        common_min = min(basket1+basket2)
        ans = 0
        
        # match minimum with maximum
        i, j = 0, len(neg) - 1
        
        while i < len(pos):
            x, y = m12[pos[i]], m21[neg[j]]
            removed = min(x, y)
            m12[pos[i]] -= removed
            m21[neg[j]] -= removed
            # either mutual swap or use the minimum pile to swap
            ans += min(min(pos[i], neg[j]) * removed, common_min * removed * 2)
            if not m12[pos[i]]:
                i += 1
            if not m21[neg[j]]:
                j -= 1
        
        return ans        
595