# N-th prime factor of a given number

Given Q queries which consist of two integers, one is number(1 <= number <= 106) and the other is N., the task is to find the N-th prime factor of the given number.

Examples:

Input: Number of Queries, Q = 4
number = 6, N = 1
number = 210, N = 3
number = 210, N = 2
number = 60, N = 2

Output:
2
5
3
3

6 has prime factors 2 and 3.
210 has prime factors 2, 3 and 6.
60 has prime factors 2 and 3.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to factorize every number and store the prime factors. Print the N-th prime factors thus stored.
Time Complexity: O(log(n)) per query.

An efficient approach is to pre-calculate all the prime factors of the number and store the numbers in a sorted order in a 2-D vector. Since the number will not be more than 106, the number of unique prime factors will be around 7-8 at max(because of 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 >= 106). Once the numbers are stored, the query can be answered in O(1) as the n-1th index will have the answer in numberth row.

Below is the implementation of the above approach:

 `// C++ program to answer queries ` `// for N-th prime factor of a number ` `#include ` `using` `namespace` `std; ` `const` `int` `N = 1000001; ` ` `  `// 2-D vector that stores prime factors ` `vector<``int``> v[N]; ` ` `  `// Function to pre-store prime ` `// factors of all numbers till 10^6 ` `void` `preprocess() ` `{ ` `    ``// calculate unique prime factors for ` `    ``// every number till 10^6 ` `    ``for` `(``int` `i = 1; i < N; i++) { ` ` `  `        ``int` `num = i; ` ` `  `        ``// find prime factors ` `        ``for` `(``int` `j = 2; j <= ``sqrt``(num); j++) { ` `            ``if` `(num % j == 0) { ` ` `  `                ``// store if prime factor ` `                ``v[i].push_back(j); ` ` `  `                ``while` `(num % j == 0) { ` `                    ``num = num / j; ` `                ``} ` `            ``} ` `        ``} ` `         `  `        ``if``(num>2) ` `        ``v[i].push_back(num); ` `         `  `    ``} ` `} ` ` `  `// Function that returns answer ` `// for every query ` `int` `query(``int` `number, ``int` `n) ` `{ ` `    ``return` `v[number][n - 1]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``// Function to pre-store unique prime factors ` `    ``preprocess(); ` ` `  `    ``// 1st query ` `    ``int` `number = 6, n = 1; ` `    ``cout << query(number, n) << endl; ` ` `  `    ``// 2nd query ` `    ``number = 210, n = 3; ` `    ``cout << query(number, n) << endl; ` ` `  `    ``// 3rd query ` `    ``number = 210, n = 2; ` `    ``cout << query(number, n) << endl; ` ` `  `    ``// 4th query ` `    ``number = 60, n = 2; ` `    ``cout << query(number, n) << endl; ` ` `  `    ``return` `0; ` `} `

Output:

```2
5
3
3
```

Time Complexity: O(1) per query and O(maxN * log(maxN)) for pre-processing, where maxN = 106.

Auxiliary Space: O(N * 8) in worst case

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