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>

class Solution:
    def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
        neg_idx = bisect_left(nums1, 0)
        neg1 = nums1[:neg_idx]
        pos1 = nums1[neg_idx:]
        neg_idx = bisect_left(nums2, 0)
        neg2 = nums2[:neg_idx]
        pos2 = nums2[neg_idx:]
        def check(m):
            res = 0
            l, r = len(neg2)-1, 0
            for x in neg1:
                while l>=0 and x*neg2[l]<=m:
                    l-=1
                res+=len(neg2)-1-l
                while r<len(pos2) and x*pos2[r]>m:
                    r+=1
                res+=len(pos2)-r
            l, r = 0, len(pos2)-1
            for i in range(len(pos1)):
                while l<len(neg2) and pos1[i]*neg2[l]<=m:
                    l+=1
                res+=l
                while r>=0 and pos1[i]*pos2[r]>m:
                    r-=1
                res+=r+1
            print(m, res)
            return res>=k
        #Binary Search
        l = min(nums1[0]*nums2[0], nums1[0]*nums2[-1], nums1[-1]*nums2[0], nums1[-1]*nums2[-1])-1
        r = max(nums1[0]*nums2[0], nums1[0]*nums2[-1], nums1[-1]*nums2[0], nums1[-1]*nums2[-1])
        while r-l>1:
            m = l+(r-l)//2
            if check(m):
                r = m
            else:
                l = m
        return r
26600