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Formatted question description: https://leetcode.ca/all/2040.html

# 2040. Kth Smallest Product of Two Sorted Arrays

• Difficulty: Hard.
• Related Topics: Array, Binary Search.
• Similar Questions: Find K Pairs with Smallest Sums, K-diff Pairs in an Array.

## 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 and nums2 are sorted.

## Solution

• class Solution {
static long inf = (long) 1e10;

public long kthSmallestProduct(int[] nums1, int[] nums2, long k) {
int n = nums2.length;
long lo = -inf - 1;
long hi = inf + 1;
while (lo < hi) {
long mid = lo + ((hi - lo) >> 1);
long cnt = 0;
for (int i : nums1) {
int l = 0;
int r = n - 1;
int p = 0;
if (0 <= i) {
while (l <= r) {
int c = l + ((r - l) >> 1);
long mul = i * (long) nums2[c];
if (mul <= mid) {
p = c + 1;
l = c + 1;
} else {
r = c - 1;
}
}
} else {
while (l <= r) {
int c = l + ((r - l) >> 1);
long mul = i * (long) nums2[c];
if (mul <= mid) {
p = n - c;
r = c - 1;
} else {
l = c + 1;
}
}
}
cnt += p;
}
if (cnt >= k) {
hi = mid;
} else {
lo = mid + 1L;
}
}
return lo;
}
}

• # 2040. Kth Smallest Product of Two Sorted Arrays
# https://leetcode.com/problems/kth-smallest-product-of-two-sorted-arrays

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 k > neg:
k -= neg
s = 1
else:
k = neg - k + 1
B1, B2 = B2, B1
s = -1

def count(A, B, t):
res = 0

j = len(B) - 1
for x in A:
while j >= 0 and x * B[j] > t:
j -= 1

res += j + 1

return res

left, right = 0, 10 ** 10

while left < right:
mid = left + (right - left) // 2

if count(A1, B1, mid) + count(A2, B2, mid) >= k:
right = mid
else:
left = mid + 1

return left * s



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).