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K-th Smallest Prime Fraction in C++
Suppose we have one sorted list, there is 1 and some prime numbers, now for every p < q in the list, we will consider fraction p/q, then we have to find which is the kth smallest fraction. We have to return an array as answer, so ans[0] will be p and ans[1] will be q.
So if the input is like [1,3,5,7], and k = 2, then the answer will be 1/5, as the fractions are 1/3, 1/5, 1/7, 3/5, 3/7, 5/7, the second smallest is 1/5.
To solve this, we will follow these steps −
- Define Data, this will take a, b and a/b
- Define an array ret of size 2
- n := size of A
- define one priority queue pq
- for initialize i := 0, when i < n, update (increase i by 1), do −
- insert Data(A[0], A[i], 0) into pq
- while K is non-zero, do −
- temp = top element of pq
- delete element from pq
- if K is same as 0, then −
- ret[0] := a of temp
- ret[1] := b of temp
- return ret
- if temp.idx + 1 < n, then −
- idx := idx of temp + 1
- insert Data(A[idx], temp.b, idx) into pq
- decrease K by 1
- return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
struct Data{
double val, a, b;
int idx;
Data(double a, double b, int c){
val = a / b;
this->a = a;
this->b = b;
idx = c;
}
};
struct Comparator{
bool operator()(Data a, Data b){
return !(a.val < b.val);
}
};
class Solution {
public:
vector<int> kthSmallestPrimeFraction(vector<int>& A, int K) {
vector <int> ret(2);
int n = A.size();
priority_queue <Data, vector <Data>, Comparator> pq;
for(int i = 0; i < n; i++){
pq.push(Data(double(A[0]), double(A[i]), 0));
}
while(K--){
Data temp = pq.top();
pq.pop();
if(K == 0){
ret[0] = temp.a;
ret[1] = temp.b;
return ret;
}
if(temp.idx + 1 < n){
int idx = temp.idx + 1;
pq.push(Data(double(A[idx]), double(temp.b), idx));
}
}
return ret;
}
};
main(){
Solution ob;
vector<int> v = {1,3,5,7};
print_vector(ob.kthSmallestPrimeFraction(v, 2));
}Input
{1,3,5,7}
2Output
[1, 5, ]
Updated on: 02-Jun-2020
300 Views
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