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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).

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