2040. Kth Smallest Product of Two Sorted Arrays
Description
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
andnums2
are sorted.
Solutions
Solution 1: Binary Search
We can use binary search to enumerate the value of the product $p$, defining the binary search interval as $[l, r]$, where $l = -\textit{max}(|\textit{nums1}[0]|, |\textit{nums1}[n - 1]|) \times \textit{max}(|\textit{nums2}[0]|, |\textit{nums2}[n - 1]|)$, $r = -l$.
For each $p$, we calculate the number of products less than or equal to $p$. If this number is greater than or equal to $k$, it means the $k$-th smallest product must be less than or equal to $p$, so we can reduce the right endpoint of the interval to $p$. Otherwise, we increase the left endpoint of the interval to $p + 1$.
The key to the problem is how to calculate the number of products less than or equal to $p$. We can enumerate each number $x$ in $\textit{nums1}$ and discuss in cases:
- If $x > 0$, then $x \times \textit{nums2}[i]$ is monotonically increasing as $i$ increases. We can use binary search to find the smallest $i$ such that $x \times \textit{nums2}[i] > p$. Then, $i$ is the number of products less than or equal to $p$, which is accumulated into the count $\textit{cnt}$;
- If $x < 0$, then $x \times \textit{nums2}[i]$ is monotonically decreasing as $i$ increases. We can use binary search to find the smallest $i$ such that $x \times \textit{nums2}[i] \leq p$. Then, $n - i$ is the number of products less than or equal to $p$, which is accumulated into the count $\textit{cnt}$;
- If $x = 0$, then $x \times \textit{nums2}[i] = 0$. If $p \geq 0$, then $n$ is the number of products less than or equal to $p$, which is accumulated into the count $\textit{cnt}$.
This way, we can find the $k$-th smallest product through binary search.
The time complexity is $O(m \times \log n \times \log M)$, where $m$ and $n$ are the lengths of $\textit{nums1}$ and $\textit{nums2}$, respectively, and $M$ is the maximum absolute value in $\textit{nums1}$ and $\textit{nums2}$.
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