2040

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] <= 105
• 1 <= k <= nums1.length * nums2.length
• nums1 and nums2 are sorted.

Similar Questions:

• Find K Pairs with Smallest Sums (Medium)
• K-diff Pairs in an Array (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/kth-smallest-product-of-two-sorted-arrays/
// Author: github.com/lzl124631x
// Time: O()
// Space: O()
class Solution {
public:
    long long kthSmallestProduct(vector<int>& A, vector<int>& B, long long k) {
        if (A.size() > B.size()) swap(A, B);
        long long L = -1e10, R = 1e10, ans = -1e10, m = A.size(), n = B.size();
        auto under = [&](long long M) {
            long long cnt = 0;
            for (int i = 0; i < A.size(); ++i) {
                auto goal = (long double)M / A[i];
                if (A[i] >= 0) { // count x < goal
                    cnt += lower_bound(begin(B), end(B), goal) - begin(B);
                } else { // count x > goal
                    cnt += n - (upper_bound(begin(B), end(B), goal) - begin(B));
                }
            }
            return cnt;
        };
        while (L < R) {
            long long M = (L + R + 1) / 2;
            if (under(M) < k) L = M;
            else R = M - 1;
        }
        return L;
    }
};