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.

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

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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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