<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:
    count1, count2 = self.return_count(nums1), self.return_count(nums2)
    sum1, sum2 = self.get_sum(count1), self.get_sum(count2)
    operations = 0
    if sum2 > sum1:
      sum1, sum2 = sum2, sum1
      count1, count2 = count2, count1
    while sum1 != sum2:
      operations += 1
      diff = sum1-sum2
      delta1 = self.closest_decrease(count1, diff)
      delta2 = self.closest_increase(count2, diff)
      if delta1[0] == delta1[1] and delta2[0] == delta2[1]:
        return -1
      if diff-abs(delta1[1]-delta1[0]) > diff-abs(delta2[1]-delta2[0]):
        count2[delta2[0]] -= 1
        count2[delta2[1]] += 1
        sum2 += abs(delta2[1]-delta2[0])
      else:
        count1[delta1[1]] -= 1
        count1[delta1[0]] += 1
        sum1 -= abs(delta1[1]-delta1[0])
    return operations

  def closest_increase(self, count, diff):
    l,r = 0,5
    for i in range(len(count)):
      if count[i] > 0:
        l = i
        break
    while r-l > diff:
      r -= 1
    return [l,r]

  def closest_decrease(self, count, diff):
    l,r = 0,0
    for i in range(len(count)-1,-1,-1):
      if count[i] > 0:
        r = i
        break
    while r-l > diff:
      l += 1
    return [l,r]

  def return_count(self, nums):
    count = [0]*6
    for num in nums:
      count[num-1] += 1
    return count

  def get_sum(self, count):
    the_sum = 0
    for i, freq, in enumerate(count):
      the_sum += (i+1)*freq
    return the_sum
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