<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:
        sqdiff = [abs(nums1[i]-nums2[i]) for i in range(len(nums1))]
        frequency = [0]*100001
        k = k1 + k2
        largest = 0
        for x in sqdiff:
            largest = max(largest,x)
            frequency[x] += 1
        
        if largest == 0:
            return 0
        while largest > 0 and k > 0:
            if frequency[largest] >= k :
                frequency[largest] -= k
                frequency[largest-1] += k
                k = 0
            else: 
                frequency[largest-1] += frequency[largest]
                k -= frequency[largest]
                frequency[largest] = 0
                largest -= 1
        sum = 0
        for i in range(len(frequency)):
            sum += frequency[i]*i**2
        return sum
                
        
        

       
954