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-1^th index will have the answer in number^th row.

Below is the implementation of the above approach:

CPP

// C++ program to answer queries 
// for N-th prime factor of a number 
#include <bits/stdc++.h> 
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; 
} 

Java

// Java program to answer queries 
// for N-th prime factor of a number 
import java.util.*; 
  
class GFG 
{ 
      
static int N = 1000001; 
  
// 2-D vector that stores prime factors 
static Vector<Integer> []v = new Vector[N]; 
  
// Function to pre-store prime 
// factors of all numbers till 10^6 
static 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 <= Math.sqrt(num); j++)  
        { 
            if (num % j == 0) 
            { 
  
                // store if prime factor 
                v[i].add(j); 
  
                while (num % j == 0)  
                { 
                    num = num / j; 
                } 
            } 
        } 
        if(num > 2) 
        v[i].add(num); 
    } 
} 
  
// Function that returns answer 
// for every query 
static int query(int number, int n) 
{ 
    return v[number].get(n - 1); 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
  
    for (int i = 0; i < N; i++) 
        v[i] = new Vector<Integer>(); 
  
    // Function to pre-store unique prime factors 
    preprocess(); 
  
    // 1st query 
    int number = 6, n = 1; 
    System.out.print(query(number, n) +"\n"); 
  
    // 2nd query 
    number = 210; n = 3; 
    System.out.print(query(number, n) +"\n"); 
  
    // 3rd query 
    number = 210; n = 2; 
    System.out.print(query(number, n) +"\n"); 
  
    // 4th query 
    number = 60; n = 2; 
    System.out.print(query(number, n) +"\n"); 
} 
} 
  
// This code is contributed by PrinciRaj1992 

Python3

# Python3 program to answer queries 
# for N-th prime factor of a number 
from math import sqrt,ceil 
N = 10001
  
# 2-D vector that stores prime factors 
v = [[] for i in range(N)] 
  
# Function to pre-store prime 
# factors of all numbers till 10^6 
def preprocess(): 
      
    # calculate unique prime factors for 
    # every number till 10^6 
    for i in range(1, N): 
  
        num = i 
  
        # find prime factors 
        for j in range(2,ceil(sqrt(num)) + 1): 
            if (num % j == 0): 
                  
                # store if prime factor 
                v[i].append(j) 
  
                while (num % j == 0): 
                    num = num // j 
  
        if(num > 2): 
            v[i].append(num) 
  
# Function that returns answer 
# for every query 
def query(number, n): 
    return v[number][n - 1] 
  
# Driver Code 
  
# Function to pre-store unique prime factors 
preprocess() 
  
# 1st query 
number = 6
n = 1
print(query(number, n)) 
  
# 2nd query 
number = 210
n = 3
print(query(number, n)) 
  
# 3rd query 
number = 210
n = 2
print(query(number, n)) 
  
# 4th query 
number = 60
n = 2
print(query(number, n)) 
  
# This code is contributed by mohit kumar 29 

C#

// C# program to answer queries 
// for N-th prime factor of a number 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
      
static int N = 100001; 
  
// 2-D vector that stores prime factors 
static List<int> []v = new List<int>[N]; 
  
// Function to pre-store prime 
// factors of all numbers till 10^6 
static 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 <= Math.Sqrt(num); j++)  
        { 
            if (num % j == 0) 
            { 
  
                // store if prime factor 
                v[i].Add(j); 
  
                while (num % j == 0)  
                { 
                    num = num / j; 
                } 
            } 
        } 
        if(num > 2) 
        v[i].Add(num); 
    } 
} 
  
// Function that returns answer 
// for every query 
static int query(int number, int n) 
{ 
    return v[number][n - 1]; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
  
    for (int i = 0; i < N; i++) 
        v[i] = new List<int>(); 
  
    // Function to pre-store unique prime factors 
    preprocess(); 
  
    // 1st query 
    int number = 6, n = 1; 
    Console.Write(query(number, n) +"\n"); 
  
    // 2nd query 
    number = 210; n = 3; 
    Console.Write(query(number, n) +"\n"); 
  
    // 3rd query 
    number = 210; n = 2;  
    Console.Write(query(number, n) +"\n"); 
  
    // 4th query 
    number = 60; n = 2; 
    Console.Write(query(number, n) +"\n"); 
} 
} 
  
// This code is contributed by PrinciRaj1992 

Output:

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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My Personal Notes

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

CPP

Java

Python3

C#

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