# 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 and nums2 are sorted.

## Solutions

• class Solution {
private int[] nums1;
private int[] nums2;

public long kthSmallestProduct(int[] nums1, int[] nums2, long k) {
this.nums1 = nums1;
this.nums2 = nums2;
int m = nums1.length;
int n = nums2.length;
int a = Math.max(Math.abs(nums1[0]), Math.abs(nums1[m - 1]));
int b = Math.max(Math.abs(nums2[0]), Math.abs(nums2[n - 1]));
long r = (long) a * b;
long l = (long) -a * b;
while (l < r) {
long mid = (l + r) >> 1;
if (count(mid) >= k) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}

private long count(long p) {
long cnt = 0;
int n = nums2.length;
for (int x : nums1) {
if (x > 0) {
int l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if ((long) x * nums2[mid] > p) {
r = mid;
} else {
l = mid + 1;
}
}
cnt += l;
} else if (x < 0) {
int l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if ((long) x * nums2[mid] <= p) {
r = mid;
} else {
l = mid + 1;
}
}
cnt += n - l;
} else if (p >= 0) {
cnt += n;
}
}
return cnt;
}
}

• class Solution {
public:
long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) {
int m = nums1.size(), n = nums2.size();
int a = max(abs(nums1[0]), abs(nums1[m - 1]));
int b = max(abs(nums2[0]), abs(nums2[n - 1]));
long long r = 1LL * a * b;
long long l = -r;
auto count = [&](long long p) {
long long cnt = 0;
for (int x : nums1) {
if (x > 0) {
int l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (1LL * x * nums2[mid] > p) {
r = mid;
} else {
l = mid + 1;
}
}
cnt += l;
} else if (x < 0) {
int l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (1LL * x * nums2[mid] <= p) {
r = mid;
} else {
l = mid + 1;
}
}
cnt += n - l;
} else if (p >= 0) {
cnt += n;
}
}
return cnt;
};
while (l < r) {
long long mid = (l + r) >> 1;
if (count(mid) >= k) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
};

• class Solution:
def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
def count(p: int) -> int:
cnt = 0
n = len(nums2)
for x in nums1:
if x > 0:
cnt += bisect_right(nums2, p / x)
elif x < 0:
cnt += n - bisect_left(nums2, p / x)
else:
cnt += n * int(p >= 0)
return cnt

mx = max(abs(nums1[0]), abs(nums1[-1])) * max(abs(nums2[0]), abs(nums2[-1]))
return bisect_left(range(-mx, mx + 1), k, key=count) - mx


• func kthSmallestProduct(nums1 []int, nums2 []int, k int64) int64 {
m := len(nums1)
n := len(nums2)
a := max(abs(nums1[0]), abs(nums1[m-1]))
b := max(abs(nums2[0]), abs(nums2[n-1]))
r := int64(a) * int64(b)
l := -r

count := func(p int64) int64 {
var cnt int64
for _, x := range nums1 {
if x > 0 {
l, r := 0, n
for l < r {
mid := (l + r) >> 1
if int64(x)*int64(nums2[mid]) > p {
r = mid
} else {
l = mid + 1
}
}
cnt += int64(l)
} else if x < 0 {
l, r := 0, n
for l < r {
mid := (l + r) >> 1
if int64(x)*int64(nums2[mid]) <= p {
r = mid
} else {
l = mid + 1
}
}
cnt += int64(n - l)
} else if p >= 0 {
cnt += int64(n)
}
}
return cnt
}

for l < r {
mid := (l + r) >> 1
if count(mid) >= k {
r = mid
} else {
l = mid + 1
}
}
return l
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}