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>

# n1 = [-2,-1,0,1,2]
# n2 = [-3,-1,2,4,5]

# A1 = [1,2], A2 = [0,1,2]
# B1 = [1,3], B2 = [2,4,5]

# neg = 2*3+2*3 = 12
# k = 12 - 3 + 1 = 10
# B1, B2 = [2,4,5], [1,3]
# -10, -8, -6, 
# 2, 4, 4, 5,  



# count(A1,B1, mid) + count(A2, B2, mid)
class Solution:
    def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
        A1, A2 = [-x for x in nums1 if x<0][::-1], [x for x in nums1 if x>=0]
        B1, B2 = [-x for x in nums2 if x<0][::-1], [x for x in nums2 if x>=0]
        neg = len(A1)*len(B2)+len(A2)*len(B1)
        if neg < k:
            k -= neg
            s = 1
        else:
            k = neg - k + 1
            B1, B2 = B2, B1
            s = -1
        def count(A, B, target):
            n1, n2 = len(A), len(B)
            j, ret = n2-1, 0
            for i in range(n1):
                while(j>=0 and A[i]*B[j] > target):
                    j -= 1
                ret += j+1
            return ret
        l, r = 0, 10**5 * 10 **5
        # print("k ----> ", k)
        while(l<r):
            mid = (l+r)//2
            chk = count(A1,B1,mid)+count(A2,B2,mid)
            # print(mid,l,r,chk)
            if chk >= k:
                r = mid
            else:
                l = mid+1
        return r*s

    


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