<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>

from collections import Counter
from heapq import heappop, heappush, heapify
from typing import List

class Solution:
    def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
        # Step 1: Calculate the absolute differences between corresponding elements of nums1 and nums2
        # and count the frequency of each difference.
        freq = Counter(abs(x1 - x2) for x1, x2 in zip(nums1, nums2) if x1 != x2)
        
        # Step 2: Create a max-heap (priority queue) from the frequency dictionary.
        # We use negative keys because Python's heapq is a min-heap by default.
        pq = [(-k, v) for k, v in freq.items()]
        heapify(pq)
        
        # Step 3: Combine the number of allowed modifications for both arrays.
        k1 += k2
        
        # Step 4: Modify the differences to minimize the sum of squared differences.
        while pq and k1:
            # Extract the largest difference from the heap.
            x, v = heappop(pq)
            
            # Check if there is another difference in the heap.
            if pq:
                xx, vv = heappop(pq)
            else:
                xx = vv = 0
            
            # Calculate the difference between the two largest differences.
            diff = xx - x
            
            # If we can reduce the current largest difference to the next largest difference
            # using the available modifications, do so.
            if diff * v <= k1:
                k1 -= diff * v
                if vv:
                    heappush(pq, (xx, v + vv))
            else:
                # Otherwise, distribute the remaining modifications to reduce the current largest difference.
                q, r = divmod(k1, v)
                k1 = 0
                heappush(pq, (x + q + 1, r))
                heappush(pq, (x + q, v - r))
                if vv:
                    heappush(pq, (xx, vv))
        
        # Step 5: Calculate the final sum of squared differences.
        return sum(x * x * v for x, v in pq)
40800