Kth Smallest Product of Two Sorted Arrays
Given two <strong>sorted 0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> as well as an integer <code>k</code>, return <em>the </em><code>k<sup>th</sup></code><em> (<strong>1-based</strong>) smallest product of </em><code>nums1[i] \* nums2[j]</code><em> where </em><code>0 &lt;= i &lt; nums1.length</code><em> and </em><code>0 &lt;= j &lt; nums2.length</code>.

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

<pre>
<strong>Input:</strong> nums1 = [2,5], nums2 = [3,4], k = 2
<strong>Output:</strong> 8
<strong>Explanation:</strong> The 2 smallest products are:
- nums1[0] * nums2[0] = 2 * 3 = 6
- nums1[0] * nums2[1] = 2 * 4 = 8
The 2<sup>nd</sup> smallest product is 8.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6
<strong>Output:</strong> 0
<strong>Explanation:</strong> The 6 smallest products are:
- nums1[0] * nums2[1] = (-4) * 4 = -16
- nums1[0] * nums2[0] = (-4) * 2 = -8
- nums1[1] * nums2[1] = (-2) * 4 = -8
- nums1[1] * nums2[0] = (-2) * 2 = -4
- nums1[2] * nums2[0] = 0 * 2 = 0
- nums1[2] * nums2[1] = 0 * 4 = 0
The 6<sup>th</sup> smallest product is 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [-2,-1,0,1,2], nums2 = [-3,-1,2,4,5], k = 3
<strong>Output:</strong> -6
<strong>Explanation:</strong> The 3 smallest products are:
- nums1[0] * nums2[4] = (-2) * 5 = -10
- nums1[0] * nums2[3] = (-2) * 4 = -8
- nums1[4] * nums2[0] = 2 * (-3) = -6
The 3<sup>rd</sup> smallest product is -6.
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums1[i], nums2[j] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums1.length * nums2.length</code></li>
	<li><code>nums1</code> and <code>nums2</code> are sorted.</li>
</ul>
Array; Binary Search
func kthSmallestProduct(nums1 []int, nums2 []int, k int64) int64 {
	m := len(nums1)
	n := len(nums2)
	a := max(abs(nums1[0]), abs(nums1[m-1]))
	b := max(abs(nums2[0]), abs(nums2[n-1]))
	r := int64(a) * int64(b)
	l := -r

	count := func(p int64) int64 {
		var cnt int64
		for _, x := range nums1 {
			if x > 0 {
				l, r := 0, n
				for l < r {
					mid := (l + r) >> 1
					if int64(x)*int64(nums2[mid]) > p {
						r = mid
					} else {
						l = mid + 1
					}
				}
				cnt += int64(l)
			} else if x < 0 {
				l, r := 0, n
				for l < r {
					mid := (l + r) >> 1
					if int64(x)*int64(nums2[mid]) <= p {
						r = mid
					} else {
						l = mid + 1
					}
				}
				cnt += int64(n - l)
			} else if p >= 0 {
				cnt += int64(n)
			}
		}
		return cnt
	}

	for l < r {
		mid := (l + r) >> 1
		if count(mid) >= k {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

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