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
impl Solution {
    pub fn kth_smallest_product(nums1: Vec<i32>, nums2: Vec<i32>, k: i64) -> i64 {
        let m = nums1.len();
        let n = nums2.len();
        let a = nums1[0].abs().max(nums1[m - 1].abs()) as i64;
        let b = nums2[0].abs().max(nums2[n - 1].abs()) as i64;
        let mut l = -a * b;
        let mut r = a * b;

        let count = |p: i64| -> i64 {
            let mut cnt = 0i64;
            for &x in &nums1 {
                if x > 0 {
                    let mut left = 0;
                    let mut right = n;
                    while left < right {
                        let mid = (left + right) / 2;
                        if (x as i64) * (nums2[mid] as i64) > p {
                            right = mid;
                        } else {
                            left = mid + 1;
                        }
                    }
                    cnt += left as i64;
                } else if x < 0 {
                    let mut left = 0;
                    let mut right = n;
                    while left < right {
                        let mid = (left + right) / 2;
                        if (x as i64) * (nums2[mid] as i64) <= p {
                            right = mid;
                        } else {
                            left = mid + 1;
                        }
                    }
                    cnt += (n - left) as i64;
                } else if p >= 0 {
                    cnt += n as i64;
                }
            }
            cnt
        };

        while l < r {
            let mid = l + (r - l) / 2;
            if count(mid) >= k {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}
