<p>You are given two arrays of integers <code>nums1</code> and <code><font face="monospace">nums2</font></code>, possibly of different lengths. The values in the arrays are between <code>1</code> and <code>6</code>, inclusive.</p>

<p>In one operation, you can change any integer&#39;s value in <strong>any </strong>of the arrays to <strong>any</strong> value between <code>1</code> and <code>6</code>, inclusive.</p>

<p>Return <em>the minimum number of operations required to make the sum of values in </em><code>nums1</code><em> equal to the sum of values in </em><code>nums2</code><em>.</em> Return <code>-1</code>​​​​​ if it is not possible to make the sum of the two arrays equal.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [<u><strong>6</strong></u>,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,<strong><u>1</u></strong>], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,<strong><u>2</u></strong>,4,5,1], nums2 = [6,1,2,2,2,2].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,1,1,1,1,1,1], nums2 = [6]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [6,6], nums2 = [1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [<strong><u>2</u></strong>,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,<strong><u>2</u></strong>], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [<strong><u>4</u></strong>].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 6</code></li>
</ul>

# class Solution:
#     def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
#         if 6*len(nums1)<len(nums2) or 6*len(nums2)<len(nums1):
#             return -1 
#         if sum(nums1)<sum(nums2):
#             nums1,nums2=nums2,nums1 
#         s1,s2=sum(nums1),sum(nums2) 
#         nums1=[1*i for i in nums1]
#         heapq.heapify(nums1)
#         heapq.heapify(nums2)
#         cnt=0 
#         while s1>s2:
#             n1,n2=nums1[0],nums2[0]
#             if -1-n1>6-n2:
#                 s1+=n1+1
#                 heapq.heapreplace(nums1,-1)
#             else:
#                 s2+= 6-n2
#                 heapq.heapreplace(nums2,6)
#             cnt+=1 
#         # return ans             

#         return cnt

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        if 6 * len(nums1) < len(nums2) or 6 * len(nums2) < len(nums1):
            return -1 
        
        s1, s2 = sum(nums1), sum(nums2)
        if s1 < s2:
            nums1, nums2 = nums2, nums1
            s1, s2 = s2, s1

        # Create max heap for nums1 and min heap for nums2
        nums1 = [-x for x in nums1]  # max heap by negating the values
        heapq.heapify(nums1)
        heapq.heapify(nums2)

        cnt = 0
        while s1 > s2:
            # Try to decrease nums1 or increase nums2 to balance sums
            if nums1 and (-nums1[0] - 1) > (6 - (nums2[0] if nums2 else 0)):
                # Reduce the largest in nums1
                largest = -heapq.heappop(nums1)
                s1 -= largest - 1
                heapq.heappush(nums1, -1)  # Replace with 1
            else:
                # Increase the smallest in nums2
                smallest = heapq.heappop(nums2)
                s2 += 6 - smallest
                heapq.heappush(nums2, 6)  # Replace with 6
            cnt += 1

        return cnt

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