<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, b1: List[int], b2: List[int]) -> int:
        minv = min(min(b1), min(b2))
        c1, c2, nums1, nums2 = Counter(b1), Counter(b2),[], []
        for k in c1:
            if (c1[k] +c2[k])%2:
                return -1
        for k in c2:
            if (c1[k] +c2[k])%2:
                return -1
        def updateNums(c1, c2, nums1, nums2):
            for k1 in c1:
                if c1[k1] > c2[k1]:
                    ca = nums1
                else:
                    ca = nums2
                for i in range(abs(c1[k1] + c2[k1] - 2 * min(c1[k1], c2[k1])) // 2):
                    ca.append(k1)
                c1[k1]= c2[k1] = 0
        updateNums(c1,c2, nums1, nums2)
        updateNums(c2,c1, nums2, nums1)
        nums1.sort()
        nums2.sort()
        n1s,n2e,n2s,n1e, other = 0, len(nums2)-1, 0, len(nums1)-1, 2*minv
        ans =0
        while n1s <= n1e and min(nums1[n1s],nums2[n2e]) < other:
            ans += min(nums1[n1s],nums2[n2e])
            n1s += 1
            n2e-=1

        while n2s <= n2e and min(nums2[n2s],nums1[n1e]) < other:
            ans += min(nums2[n2s],nums1[n1e])
            n2s += 1
            n1e-=1
        ans  +=  other *(n2e-n2s+1)
    



        return ans




        
        
704