- Difficulty: Hard
- Tags: LeetCode, Hard, Array, Binary Search, leetcode-2040, Two Pointers, O((m + n) * logr), O(1)
Problem
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.