LeetCode-in-Java
2040. Kth Smallest Product of Two Sorted Arrays
Hard
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] <= 1051 <= k <= nums1.length * nums2.lengthnums1andnums2are sorted.
Solution
public 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;
}
}