<p>You are given two positive <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, both of length <code>n</code>.</p>

<p>The <strong>sum of squared difference</strong> of arrays <code>nums1</code> and <code>nums2</code> is defined as the <strong>sum</strong> of <code>(nums1[i] - nums2[i])<sup>2</sup></code> for each <code>0 &lt;= i &lt; n</code>.</p>

<p>You are also given two positive integers <code>k1</code> and <code>k2</code>. You can modify any of the elements of <code>nums1</code> by <code>+1</code> or <code>-1</code> at most <code>k1</code> times. Similarly, you can modify any of the elements of <code>nums2</code> by <code>+1</code> or <code>-1</code> at most <code>k2</code> times.</p>

<p>Return <em>the minimum <strong>sum of squared difference</strong> after modifying array </em><code>nums1</code><em> at most </em><code>k1</code><em> times and modifying array </em><code>nums2</code><em> at most </em><code>k2</code><em> times</em>.</p>

<p><strong>Note</strong>: You are allowed to modify the array elements to become <strong>negative</strong> integers.</p>

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

<pre>
<strong>Input:</strong> nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
<strong>Output:</strong> 579
<strong>Explanation:</strong> The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0. 
The sum of square difference will be: (1 - 2)<sup>2 </sup>+ (2 - 10)<sup>2 </sup>+ (3 - 20)<sup>2 </sup>+ (4 - 19)<sup>2</sup>&nbsp;= 579.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
<strong>Output:</strong> 43
<strong>Explanation:</strong> One way to obtain the minimum sum of square difference is: 
- Increase nums1[0] once.
- Increase nums2[2] once.
The minimum of the sum of square difference will be: 
(2 - 5)<sup>2 </sup>+ (4 - 8)<sup>2 </sup>+ (10 - 7)<sup>2 </sup>+ (12 - 9)<sup>2</sup>&nbsp;= 43.
Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.</pre>

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

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= k1, k2 &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
        diffs = []
        for i in range(0, len(nums1)):
            stuff = (-abs(nums1[i]-nums2[i]), 1)
            heapq.heappush(diffs, stuff)
        k = k1 + k2
        while k > 0:
            #print("{} left => {}".format(k, diffs))
            cur_biggest, mult = heapq.heappop(diffs)
            while len(diffs) > 0 and diffs[0][0] == cur_biggest:
                lol, extra = heapq.heappop(diffs)
                mult += extra
            cur_biggest = -cur_biggest
            if cur_biggest == 0:
                break
            to_remove = cur_biggest
            if len(diffs) > 0:
                to_remove += diffs[0][0]
            #to_remove = cur_biggest
            #if next_biggest is None:
            #    rem_all_amount = k // mult
            #    extra_removes = k % mult
            #    nonextra_count = (mult - extra_removes)
            #    heapq.heappush(diffs, (-(cur_biggest - rem_all_amount - 1), extra_removes))
            #    heapq.heappush(diffs, (-(cur_biggest - rem_all_amount), nonextra_count))
            #    k = 0
            #else:
            #    to_remove = cur_biggest - next_biggest
            if to_remove * mult <= k: # can remove everything
                k -= to_remove * mult
                heapq.heappush(diffs, (-cur_biggest+to_remove, mult))
            else: # deal with end
                rem_all_amount = k // mult
                extra_removes = k % mult
                nonextra_count = (mult - extra_removes)
                heapq.heappush(diffs, (-(cur_biggest - rem_all_amount - 1), extra_removes))
                heapq.heappush(diffs, (-(cur_biggest - rem_all_amount), nonextra_count))
                k = 0
        result = 0
        for diff, mult in diffs:
            result += mult * (diff**2)
        return result
990