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N-th prime factor of a given number

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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

Explanations:

  • For number = 6, The prime factors 2 and 3. So, 2 is the 1st one.
  • For number = 210, The prime factors 2, 3, 5 and 7. So, 5 is the 3rd one.
  • For number = 210, The prime factors 2, 3, 5 and 7. So, 3 is the 2nd one.
  • For number = 60, The prime factors 2, 2, 3 and 5. So, 2 is the 2nd one.
     

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: 

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


Javascript




<script>
// Javascript program to answer queries
// for N-th prime factor of a number
const N = 1000001;
 
// 2-D vector that stores prime factors
let v = new Array();
 
for (let i = 0; i < N; i++) {
    v.push(new Array())
}
// Function to pre-store prime
// factors of all numbers till 10^6
function preprocess()
{
 
    // calculate unique prime factors for
    // every number till 10^6
    for (let i = 1; i < N; i++) {
 
        let num = i;
 
        // find prime factors
        for (let j = 2; j <= Math.sqrt(num); j++) {
            if (num % j == 0) {
 
                // store if prime factor
                v[i].push(j);
 
                while (num % j == 0) {
                    num = num / j;
                }
            }
        }
 
        if (num > 2)
            v[i].push(num);
 
    }
}
 
// Function that returns answer
// for every query
function query(number, n) {
    return v[number][n - 1];
}
 
// Driver Code
 
// Function to pre-store unique prime factors
preprocess();
 
// 1st query
let number = 6, n = 1;
document.write(query(number, n) + "<br>");
 
// 2nd query
number = 210, n = 3;
document.write(query(number, n) + "<br>");
 
// 3rd query
number = 210, n = 2;
document.write(query(number, n) + "<br>");
 
// 4th query
number = 60, n = 2;
document.write(query(number, n) + "<br>");
 
// This code is contributed gfgking
</script>


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

Another Efficient Approach:

Another efficient approach is to use Sieve of Eratosthenes. Note that this solution is efficient when we need k-th prime factor for multiple test cases.

Approach:

The approach is to do preprocessing and store least prime factor of all numbers in given range. Once we have least prime factors stored in an array, we can find k-th prime factor by repeatedly dividing n with least prime factor while it is divisible, then repeating the process for reduced n.

Steps:

  1. The program starts by defining a function called “isPrime” that checks if a given number is prime or not. It returns a boolean value true if the number is prime and false otherwise.
  2. The program then defines a function called “AllPrimeFactors” that takes an integer argument N and returns a vector of all possible prime factors of the number. It uses a loop to iterate through all numbers from 2 to the square root of N and checks if each number is a factor of N. If it is, it adds the number to the vector and divides N by that number. It continues this process until N is no longer divisible by the current factor. If the remaining N is also a prime number, it adds it to the vector.
  3. The program defines a function called “query” that takes two integer arguments – a number and an integer k. It first calls the “AllPrimeFactors” function to get all possible prime factors of the number. It then checks if the number itself is a prime number and if the value of k is 1. If both conditions are true, it returns the number as the answer to the query. If the size of the vector of prime factors is less than k, it returns -1. Otherwise, it returns the k-th element of the vector as the answer to the query.
  4. The program then defines the “main” function which is the entry point of the program. It creates four queries by setting different values of “number” and “n”. It calls the “query” function for each query and prints the result to the console.
  5. Finally, the program ends by returning 0 from the main function.

C++




// C++ program to answer queries
// for N-th prime factor of a number
// Using Sieve of Eratosthenes method
#include <bits/stdc++.h>
using namespace std;
 
// Check if the number if prime or not
bool isPrime(int n)
{
    if (n <= 1) {
        return false;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}
 
// Function to find all the possible prime factors
vector<int> AllPrimeFactors(int N)
{
    vector<int> v;
    for (int i = 2; i * i <= N; i++) {
        while (N % i == 0) {
            v.push_back(i);
            N = N / i;
            if (isPrime(N)) {
                v.push_back(N);
                N = N / N;
            }
        }
    }
    return v;
}
 
// Function that returns answer
// for every query
int query(int n, int k)
{
    vector<int> ans = AllPrimeFactors(n);
    if (isPrime(n) && k == 1) {
        return n;
    }
    if (ans.size() < k) {
        return -1;
    }
    return ans[k - 1];
}
 
// Driver Code
int main()
{
    // 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;
}
 
// This code is contributed by Susobhan Akhuli


Java




// Java program to answer queries
// for N-th prime factor of a number
// Using Sieve of Eratosthenes method
import java.util.Vector;
 
public class GFG {
 
    // Check if the number if prime or not
    public static boolean isPrime(int n)
    {
        if (n <= 1) {
            return false;
        }
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }
 
    // Function to find all the possible prime factors
    public static Vector<Integer> AllPrimeFactors(int N)
    {
        Vector<Integer> v = new Vector<Integer>();
        for (int i = 2; i * i <= N; i++) {
            while (N % i == 0) {
                v.add(i);
                N = N / i;
                if (isPrime(N)) {
                    v.add(N);
                    N = N / N;
                }
            }
        }
        return v;
    }
 
    // Function that returns answer
    // for every query
    public static int query(int n, int k)
    {
        Vector<Integer> ans = AllPrimeFactors(n);
        if (isPrime(n) && k == 1) {
            return n;
        }
        if (ans.size() < k) {
            return -1;
        }
        return ans.get(k - 1);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // 1st query
        int number = 6, n = 1;
        System.out.println(query(number, n));
 
        // 2nd query
        number = 210;
        n = 3;
        System.out.println(query(number, n));
 
        // 3rd query
        number = 210;
        n = 2;
        System.out.println(query(number, n));
 
        // 4th query
        number = 60;
        n = 2;
        System.out.println(query(number, n));
    }
}
 
// This code is contributed by Susobhan Akhuli


Python3




# Python program to answer queries
# for N-th prime factor of a number
# Using Sieve of Eratosthenes method
import math
 
# Check if the number is prime or not
def isPrime(n):
    if n <= 1:
        return False
    for i in range(2, int(math.sqrt(n))+1):
        if n % i == 0:
            return False
    return True
 
# Function to find all the possible prime factors
def AllPrimeFactors(N):
    v = []
    i = 2
    while i * i <= N:
        while N % i == 0:
            v.append(i)
            N = N // i
            if isPrime(N):
                v.append(N)
                N = N // N
        i += 1
    if N > 1:
        v.append(N)
    return v
 
# Function that returns answer for every query
def query(n, k):
    ans = AllPrimeFactors(n)
    if isPrime(n) and k == 1:
        return n
    if len(ans) < k:
        return -1
    return ans[k-1]
 
# Driver Code
if __name__ == '__main__':
    # 1st query
    number, n = 6, 1
    print(query(number, n))
 
    # 2nd query
    number, n = 210, 3
    print(query(number, n))
 
    # 3rd query
    number, n = 210, 2
    print(query(number, n))
 
    # 4th query
    number, n = 60, 2
    print(query(number, n))
 
# This code is contributed by Susobhan Akhuli


C#




// C# program to answer queries
// for N-th prime factor of a number
// Using Sieve of Eratosthenes method
using System;
using System.Collections.Generic;
 
class Program
{
 
  // Check if the number if prime or not
  static bool IsPrime(int n)
  {
    if (n <= 1) {
      return false;
    }
    for (int i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        return false;
      }
    }
    return true;
  }
 
  // Function to find all the possible prime factors
  static List<int> AllPrimeFactors(int N)
  {
    List<int> v = new List<int>();
    for (int i = 2; i * i <= N; i++) {
      while (N % i == 0) {
        v.Add(i);
        N = N / i;
        if (IsPrime(N)) {
          v.Add(N);
          N = N / N;
        }
      }
    }
    return v;
  }
 
  // Function that returns answer
  // for every query
  static int Query(int n, int k)
  {
    List<int> ans = AllPrimeFactors(n);
    if (IsPrime(n) && k == 1) {
      return n;
    }
    if (ans.Count < k) {
      return -1;
    }
    return ans[k - 1];
  }
 
  // Driver Code
  static void Main(string[] args)
  {
    // 1st query
    int number = 6, n = 1;
    Console.WriteLine(Query(number, n));
 
    // 2nd query
    number = 210;
    n = 3;
    Console.WriteLine(Query(number, n));
 
    // 3rd query
    number = 210;
    n = 2;
    Console.WriteLine(Query(number, n));
 
    // 4th query
    number = 60;
    n = 2;
    Console.WriteLine(Query(number, n));
  }
}
 
// This code is contributed by Susobhan Akhuli


Javascript




function isPrime(n) {
  if (n <= 1) {
    return false;
  }
  for (let i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
 
function allPrimeFactors(N) {
  let v = [];
  for (let i = 2; i * i <= N; i++) {
    while (N % i == 0) {
      v.push(i);
      N = N / i;
      if (isPrime(N)) {
        v.push(N);
        N = N / N;
      }
    }
  }
  return v;
}
 
function query(n, k) {
  let ans = allPrimeFactors(n);
  if (isPrime(n) && k == 1) {
    return n;
  }
  if (ans.length < k) {
    return -1;
  }
  return ans[k - 1];
}
 
// Test cases
let number = 6, n = 1;
console.log(query(number, n));
 
number = 210, n = 3;
console.log(query(number, n));
 
number = 210, n = 2;
console.log(query(number, n));
 
number = 60, n = 2;
console.log(query(number, n));


Output

2
5
3
2

Time Complexity: O(N*log(log(N))), where N is the number.
Auxiliary Space: O(N)



Last Updated : 26 Apr, 2023
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