minimum-sum-of-squared-difference

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Algorithms

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

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

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.

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

Note: You are allowed to modify the array elements to become negative integers.

&nbsp;
Example 1:

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

Example 2:

<pre>
Input: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
Output: 43
Explanation: 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)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2&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>

&nbsp;
Constraints:

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 105</code></li>
	<li><code>0 &lt;= k1, k2 &lt;= 109</code></li>
</ul>