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

May LeetCoding Challenge 10

Description

K-th Smallest Prime Fraction

You are given a sorted integer array arr containing 1 and prime numbers, where all the integers of arr are unique. You are also given an integer k.

For every i and j where 0 <= i < j < arr.length, we consider the fraction arr[i] / arr[j].

Return the kth smallest fraction considered. Return your answer as an array of integers of size 2, where answer[0] == arr[i] and answer[1] == arr[j].

 

Example 1:

Input: arr = [1,2,3,5], k = 3
Output: [2,5]
Explanation: The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.
The third fraction is 2/5.

Example 2:

Input: arr = [1,7], k = 1
Output: [1,7]

 

Constraints:

• 2 <= arr.length <= 1000
• 1 <= arr[i] <= 3 * 104
• arr[0] == 1
• arr[i] is a prime number for i > 0.
• All the numbers of arr are unique and sorted in strictly increasing order.
• 1 <= k <= arr.length * (arr.length - 1) / 2

 

Follow up: Can you solve the problem with better than O(n2) complexity?

Solution

struct s{
  double val;
  int up;
  int dn;
  
  bool operator<(const s& b){
    return val<b.val;
  }
}v[2000000];
class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& A, int K) {
    vector<int> ans;
    int cnt = 0;
    for(int i=1;i<A.size();++i){
      for(int j=0;j<i;++j){
        v[cnt].val = 1.0*A[j]/A[i];
        v[cnt].up = A[j];
        v[cnt].dn = A[i];
        ++cnt;
      }
    }
    nth_element(v,v+K-1,v+cnt);
    // for(int i=0;i<v.size();++i){
    //     cout<<v[i].up<<" "<<v[i].dn<<endl;
    // }
    ans.push_back(v[K-1].up);
    ans.push_back(v[K-1].dn);
    return ans;
  }
};