2025-06-25 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 25

Description

Kth Smallest Product of Two Sorted Arrays

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 {
  long long countLE(const vector<int>& nums1, const vector<int>& nums2, long long target) {
    long long result = 0;
    int len2 = nums2.size();
    for(auto n1 : nums1) {
      if(!n1) {
        if(target >= 0) result += len2;
        continue;
      }
      int low = 0;
      int high = len2 - 1;
      while(low <= high) {
        int mid = (low + high) / 2;
        if((n1 > 0 && 1LL * n1 * nums2[mid] <= target) ||
           (n1 < 0 && 1LL * n1 * nums2[mid] > target)) {
          low = mid + 1;
        } else {
          high = mid - 1;
        }
      }
      result += n1 < 0 ? len2 - low : low;
    }
    return result;
  }
public:
  long long kthSmallestProduct(vector<int>& nums1, vector<int>& nums2, long long k) {
    long long mmin = min({1LL * nums1.front() * nums2.back(), 1LL * nums1.back() * nums2.front(), 1LL * nums1.front() * nums2.front(), 1LL * nums2.back() * nums2.back()});
    long long mmax = max({1LL * nums1.front() * nums2.back(), 1LL * nums1.back() * nums2.front(), 1LL * nums1.front() * nums2.front(), 1LL * nums2.back() * nums2.back()});

    // cout << "~~~~~~~~~~~~~~~~~~~~~" << endl;
    while(mmin < mmax) {
      long long mmid = (mmin + mmax - (mmax <= 0)) / 2;
      // cout << mmin << ' ' << mmax << ' ' << mmid << ' ' << countLE(nums1, nums2, mmid) << endl;
      if(countLE(nums1, nums2, mmid) < k) {
        mmin = mmid + 1;
      } else {
        mmax = mmid;
      }
    }
    return mmin;
  }
};

// Accepted
// 112/112 cases passed (541 ms)
// Your runtime beats 71.3 % of cpp submissions
// Your memory usage beats 41.69 % of cpp submissions (98.1 MB)
Algorithm