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>

from typing import List

class Solution:
    def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
        # Separate negative and non-negative elements in nums1
        neg = [x for x in nums1 if x < 0]
        pos = [x for x in nums1 if x >= 0]

        def countProductsLessThanOrEqual(val: int) -> int:
            """Returns the count of products that are <= val."""
            count = 0
            lo, hi = 0, len(nums2) - 1

            # Iterate through negative elements followed by positive elements
            if val >= 0:
                candidates = neg[::-1] + pos
            else:
                candidates = neg + pos[::-1]

            for x in candidates:
                if x < 0:
                    # For negative x, we search for nums2[j] such that x * nums2[j] <= val
                    while lo < len(nums2) and x * nums2[lo] > val:
                        lo += 1
                    count += len(nums2) - lo
                elif x == 0:
                    # For x = 0, all products are 0, which are <= val if val >= 0
                    if val >= 0:
                        count += len(nums2)
                else:
                    # For positive x, we search for nums2[j] such that x * nums2[j] <= val
                    while hi >= 0 and x * nums2[hi] > val:
                        hi -= 1
                    count += hi + 1

            return count

        # Binary search for the k-th smallest product
        low, high = -10**10, 10**10 + 1
        while low < high:
            mid = (low + high) // 2
            if countProductsLessThanOrEqual(mid) < k:
                low = mid + 1
            else:
                high = mid
        
        return low
26000