# Solving Leetcode Interviews in Seconds with AI: Kth Smallest Product of Two Sorted Arrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2040" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given two sorted 0-indexed integer arrays nums1 and nums2 as well as an integer k, return the kth (1-based) smallest product of nums1[i] * nums2[j] where 0 <= i < nums1.length and 0 <= j < nums2.length.   Example 1:  Input: nums1 = [2,5], nums2 = [3,4], k = 2 Output: 8 Explanation: The 2 smallest products are: - nums1[0] * nums2[0] = 2 * 3 = 6 - nums1[0] * nums2[1] = 2 * 4 = 8 The 2nd smallest product is 8.  Example 2:  Input: nums1 = [-4,-2,0,3], nums2 = [2,4], k = 6 Output: 0 Explanation: 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 6th smallest product is 0.  Example 3:  Input: nums1 = [-2,-1,0,1,2], nums2 = [-3,-1,2,4,5], k = 3 Output: -6 Explanation: 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 3rd smallest product is -6.    Constraints:  1 <= nums1.length, nums2.length <= 5 * 104 -105 <= nums1[i], nums2[j] <= 105 1 <= k <= nums1.length * nums2.length nums1 and nums2 are sorted.  

	# Explanation
	Here's the solution:

*   **Binary Search for the Kth Smallest Product:** The core idea is to use binary search on the range of possible product values to find the kth smallest product. We determine a `mid` value in the range and efficiently count how many products are less than or equal to `mid`.
*   **Efficient Counting of Products <= mid:** For each number in `nums1`, we use binary search in `nums2` to find how many numbers in `nums2` when multiplied by the current number in `nums1` result in a product less than or equal to `mid`.  Care is taken to handle negative, zero, and positive numbers correctly.

*   **Time Complexity:** O(m log(m\*n) log(range)), where `m` and `n` are the lengths of `nums1` and `nums2` and range is the range of possible products.  Space Complexity: O(1)

	
	# Code
	```python
	def kthSmallestProduct(nums1, nums2, k):
    def count_less_equal(mid):
        count = 0
        for num1 in nums1:
            low = 0
            high = len(nums2) - 1
            idx = -1
            while low <= high:
                m = (low + high) // 2
                if num1 * nums2[m] <= mid:
                    idx = m
                    low = m + 1
                else:
                    high = m - 1
            if idx != -1:
                count += (idx + 1)
        return count

    low = -10**10
    high = 10**10
    ans = 0
    while low <= high:
        mid = (low + high) // 2
        if count_less_equal(mid) >= k:
            ans = mid
            high = mid - 1
        else:
            low = mid + 1
    return ans
	```