Minimum Sum of Squared Difference
<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>
Greedy; Array; Binary Search; Sorting; Heap (Priority Queue)
func minSumSquareDiff(nums1 []int, nums2 []int, k1 int, k2 int) int64 {
	k := k1 + k2
	s, mx := 0, 0
	n := len(nums1)
	d := make([]int, n)
	for i, v := range nums1 {
		d[i] = abs(v - nums2[i])
		s += d[i]
		mx = max(mx, d[i])
	}
	if s <= k {
		return 0
	}
	left, right := 0, mx
	for left < right {
		mid := (left + right) >> 1
		t := 0
		for _, v := range d {
			t += max(v-mid, 0)
		}
		if t <= k {
			right = mid
		} else {
			left = mid + 1
		}
	}
	for i, v := range d {
		k -= max(v-left, 0)
		d[i] = min(v, left)
	}
	for i, v := range d {
		if k <= 0 {
			break
		}
		if v == left {
			d[i]--
			k--
		}
	}
	ans := 0
	for _, v := range d {
		ans += v * v
	}
	return int64(ans)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}