Smallest integer which has n factors or more

Given n, find the smallest integer which has n factors or more. It may be assumed that the result is less than 1000001.

Examples:

Input : n = 3
Output : 4
Explanation: 4 has factors 1, 2 and 4.

Input : n = 2 
Output : 2
Explanation: 2 has one factor 1 and 2.

There are many methods to calculate the number of factors, but the efficient one can be found here

Simple Approach: A simple approach will be to run a loop to find out the factors of a number. One for finding out the factors of a number in O(x) is to run a loop from 1 to x and see all numbers that divide x.
Time Complexity: O(x) for every number x that we try until we find answer or reach limit.

Efficient Approach:We can find out factors in sqrt(x)for every iteration.
Time Complexity: O(sqrt(x)) for every number x that we try until we find answer or reach limit.

Best Approach will be to traverse for every number and calculate the number of factors. Then check if the count is equal to or more then n then we get our desired smallest integer with n or more factors.

Below is the implementation of the above approach:

C++

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// c++ program to print the smallest
// integer with n factors or more
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 1000001;
  
// array to store prime factors
int factor[MAX] = { 0 };
  
// function to generate all prime factors
// of numbers from 1 to 10^6
void generatePrimeFactors()
{
    factor[1] = 1;
  
    // Initializes all the positions
    // with their value.
    for (int i = 2; i < MAX; i++)
        factor[i] = i;
  
    // Initializes all multiples of 2 with 2
    for (int i = 4; i < MAX; i += 2)
        factor[i] = 2;
  
    // A modified version of Sieve of
    // Eratosthenes to store the smallest
    // prime factor that divides every number.
    for (int i = 3; i * i < MAX; i++) {
  
        // check if it has no prime factor.
        if (factor[i] == i) {
  
            // Initializes of j starting from i*i
            for (int j = i * i; j < MAX; j += i) {
  
                // if it has no prime factor
                // before, then stores the
                // smallest prime divisor
                if (factor[j] == j)
                    factor[j] = i;
            }
        }
    }
}
  
// function to calculate number of factors
int calculateNoOFactors(int n)
{
    if (n == 1)
        return 1;
  
    int ans = 1;
  
    // stores the smallest prime number
    // that divides n
    int dup = factor[n];
  
    // stores the count of number of times
    // a prime number divides n.
    int c = 1;
  
    // reduces to the next number after prime
    // factorization of n
    int j = n / factor[n];
  
    // false when prime factorization is done
    while (j != 1) {
  
        // if the same prime number is dividing n,
        // then we increase the count
        if (factor[j] == dup)
            c += 1;
  
        /* if its a new prime factor that is
           factorizing n, then we again set c=1 
           and change dup to the new prime factor,
           and apply the formula explained above. */
        else {
            dup = factor[j];
            ans = ans * (c + 1);
            c = 1;
        }
  
        // prime factorizes a number
        j = j / factor[j];
    }
  
    // for the last prime factor
    ans = ans * (c + 1);
  
    return ans;
}
  
// function to find the smallest integer
// with n factors or more.
int smallest(int n)
{
    for (int i = 1;; i++)
  
        // check if no of factors is more
        // than n or not
        if (calculateNoOFactors(i) >= n)
            return i;
}
  
// Driver program to test above function
int main()
{
    // generate prime factors of number
    // upto 10^6
    generatePrimeFactors();
  
    int n = 4;
    cout << smallest(n);
    return 0;
}

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Java

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// Java program to print the smallest
// integer with n factors or more
import java.util.*;
import java.lang.*;
  
public class GfG{
    private static final int MAX = 1000001;
  
    // array to store prime factors
    private static final int[] factor = new int [MAX];
  
    // function to generate all prime factors
    // of numbers from 1 to 10^6
    public static void generatePrimeFactors()
    {
        factor[1] = 1;
  
        // Initializes all the positions
        // with their value.
        for (int i = 2; i < MAX; i++)
            factor[i] = i;
  
        // Initializes all multiples of 2 with 2
        for (int i = 4; i < MAX; i += 2)
            factor[i] = 2;
  
        // A modified version of Sieve of
        // Eratosthenes to store the smallest
        // prime factor that divides every number.
        for (int i = 3; i * i < MAX; i++) {
  
            // check if it has no prime factor.
            if (factor[i] == i) {
  
                // Initializes of j starting from i*i
                for (int j = i * i; j < MAX; j += i) {
  
                    // if it has no prime factor
                    // before, then stores the
                    // smallest prime divisor
                    if (factor[j] == j)
                        factor[j] = i;
                }
            }
        }
    }
  
    // function to calculate number of factors
    public static int calculateNoOFactors(int n)
    {
        if (n == 1)
            return 1;
  
        int ans = 1;
  
        // stores the smallest prime number
        // that divides n
        int dup = factor[n];
  
        // stores the count of number of times
        // a prime number divides n.
        int c = 1;
  
        // reduces to the next number after prime
        // factorization of n
        int j = n / factor[n];
  
        // false when prime factorization is done
        while (j != 1) {
  
            // if the same prime number is dividing n,
            // then we increase the count
            if (factor[j] == dup)
                c += 1;
  
            /* if its a new prime factor that is
            factorizing n, then we again set c=1 
            and change dup to the new prime factor,
            and apply the formula explained above. */
            else {
                dup = factor[j];
                ans = ans * (c + 1);
                c = 1;
            }
  
            // prime factorizes a number
            j = j / factor[j];
        }
  
        // for the last prime factor
        ans = ans * (c + 1);
  
        return ans; 
    }
  
    // function to find the smallest integer
    // with n factors or more.
    public static int smallest(int n)
    {
        for (int i = 1;; i++)
  
            // check if no of factors is more
            // than n or not
            if (calculateNoOFactors(i) >= n)
                return i;
    }
      
    // driver function
    public static void main(String args[])
    {
        // generate prime factors of number
        // upto 10^6
        generatePrimeFactors();
  
        int n = 4;
        System.out.println(smallest(n));
    }
}
  
/* This code is contributed by Sagar Shukla */

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Python3

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# Python3 program to print the 
# smallest integer with n
# factors or more
MAX = 100001;
  
# array to store
# prime factors
factor = [0] * MAX;
  
# function to generate all 
# prime factors of numbers
# from 1 to 10^6
def generatePrimeFactors():
      
    factor[1] = 1;
  
    # Initializes all the 
    # positions with their value.
    for i in range(2, MAX):
        factor[i] = i;
  
    # Initializes all
    # multiples of 2 with 2
    i = 4
    while(i < MAX):
        factor[i] = 2;
        i += 2;
  
    # A modified version of 
    # Sieve of Eratosthenes 
    # to store the smallest
    # prime factor that 
    # divides every number.
    i = 3
    while(i * i < MAX):
  
        # check if it has
        # no prime factor.
        if (factor[i] == i):
  
            # Initializes of j 
            # starting from i*i
            j = i * i;
            while(j < MAX): 
  
                # if it has no prime factor
                # before, then stores the
                # smallest prime divisor
                if (factor[j] == j):
                    factor[j] = i;
                j += i;
        i += 1;
  
# function to calculate 
# number of factors
def calculateNoOFactors(n):
    if (n == 1):
        return 1;
  
    ans = 1;
  
    # stores the smallest prime 
    # number that divides n
    dup = factor[n];
  
    # stores the count of 
    # number of times a 
    # prime number divides n.
    c = 1;
  
    # reduces to the next
    # number after prime
    # factorization of n
    j = int(n / factor[n]);
  
    # false when prime 
    # factorization is done
    while (j != 1):
  
        # if the same prime number 
        # is dividing n, then we 
        # increase the count
        if (factor[j] == dup):
            c += 1;
  
        # if its a new prime factor 
        # that is factorizing n, then 
        # we again set c=1 and change 
        # dup to the new prime factor,
        # and apply the formula 
        # explained above. 
        else:
            dup = factor[j];
            ans = ans * (c + 1);
            c = 1;
  
        # prime factorizes a number
        j = int(j / factor[j]);
  
    # for the last prime factor
    ans = ans * (c + 1);
  
    return ans;
  
# function to find the 
# smallest integer with
# n factors or more.
def smallest(n):
    i = 1;
    while(True):
          
        # check if no of 
        # factors is more
        # than n or not
        if (calculateNoOFactors(i) >= n):
            return i;
        i += 1;
  
# Driver Code
  
# generate prime factors 
# of number upto 10^6
generatePrimeFactors();
  
n = 4;
print(smallest(n));
  
# This code is contributed by mits

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C#

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// C# program to print the smallest
// integer with n factors or more
using System;
  
class GfG {
    private static int MAX = 1000001;
  
    // array to store prime factors
    private static int []factor = new int [MAX];
  
    // function to generate all prime 
    // factorsof numbers from 1 to 10^6
    public static void generatePrimeFactors()
    {
        factor[1] = 1;
  
        // Initializes all the positions
        // with their value.
        for (int i = 2; i < MAX; i++)
            factor[i] = i;
  
        // Initializes all multiples of 2 with 2
        for (int i = 4; i < MAX; i += 2)
            factor[i] = 2;
  
        // A modified version of Sieve of
        // Eratosthenes to store the smallest
        // prime factor that divides every number.
        for (int i = 3; i * i < MAX; i++) {
  
            // check if it has no prime factor.
            if (factor[i] == i) {
  
                // Initializes of j starting from i*i
                for (int j = i * i; j < MAX; j += i)
                {
  
                    // if it has no prime factor
                    // before, then stores the
                    // smallest prime divisor
                    if (factor[j] == j)
                        factor[j] = i;
                }
            }
        }
    }
  
    // function to calculate number of factors
    public static int calculateNoOFactors(int n)
    {
        if (n == 1)
            return 1;
  
        int ans = 1;
  
        // stores the smallest prime number
        // that divides n
        int dup = factor[n];
  
        // stores the count of number of times
        // a prime number divides n.
        int c = 1;
  
        // reduces to the next number after prime
        // factorization of n
        int j = n / factor[n];
  
        // false when prime factorization is done
        while (j != 1) {
  
            // if the same prime number is dividing n,
            // then we increase the count
            if (factor[j] == dup)
                c += 1;
  
            // if its a new prime factor that is
            // factorizing n, then we again set c=1 
            // and change dup to the new prime factor,
            // and apply the formula explained above.
            else {
                dup = factor[j];
                ans = ans * (c + 1);
                c = 1;
            }
  
            // prime factorizes a number
            j = j / factor[j];
        }
  
        // for the last prime factor
        ans = ans * (c + 1);
  
        return ans; 
    }
  
    // function to find the smallest integer
    // with n factors or more.
    public static int smallest(int n)
    {
        for (int i = 1; ; i++)
  
            // check if no of factors is more
            // than n or not
            if (calculateNoOFactors(i) >= n)
                return i;
    }
      
    // Driver Code
    public static void Main()
    {
          
        // generate prime factors of 
        // number upto 10^6
        generatePrimeFactors();
  
        int n = 4;
        Console.Write(smallest(n));
    }
}
  
// This code is contributed by Nitin Mittal.

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PHP

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<?php
// PHP program to print the 
// smallest integer with n
// factors or more
  
$MAX = 100001;
  
// array to store
// prime factors
$factor = array_fill(0, $MAX, 0);
  
// function to generate all 
// prime factors of numbers
// from 1 to 10^6
function generatePrimeFactors()
{
    global $MAX;
    global $factor;
    $factor[1] = 1;
  
    // Initializes all the 
    // positions with their value.
    for ($i = 2; $i < $MAX; $i++)
        $factor[$i] = $i;
  
    // Initializes all
    // multiples of 2 with 2
    for ($i = 4; $i < $MAX; $i += 2)
        $factor[$i] = 2;
  
    // A modified version of 
    // Sieve of Eratosthenes 
    // to store the smallest
    // prime factor that 
    // divides every number.
    for ($i = 3; $i * $i < $MAX; $i++)
    {
  
        // check if it has
        // no prime factor.
        if ($factor[$i] == $i
        {
  
            // Initializes of j 
            // starting from i*i
            for ($j = $i * $i
                 $j < $MAX; $j += $i
            {
  
                // if it has no prime factor
                // before, then stores the
                // smallest prime divisor
                if ($factor[$j] == $j)
                    $factor[$j] = $i;
            }
        }
    }
}
  
// function to calculate 
// number of factors
function calculateNoOFactors($n)
{
    global $factor;
    if ($n == 1)
        return 1;
  
    $ans = 1;
  
    // stores the smallest prime 
    // number that divides n
    $dup = $factor[$n];
  
    // stores the count of 
    // number of times a 
    // prime number divides n.
    $c = 1;
  
    // reduces to the next
    // number after prime
    // factorization of n
    $j = (int)($n / $factor[$n]);
  
    // false when prime 
    // factorization is done
    while ($j != 1) 
    {
  
        // if the same prime number 
        // is dividing n, then we 
        // increase the count
        if ($factor[$j] == $dup)
            $c += 1;
  
        /* if its a new prime factor 
        that is factorizing n, then 
        we again set c=1 and change 
        dup to the new prime factor,
        and apply the formula 
        explained above. */
        else 
        {
            $dup = $factor[$j];
            $ans = $ans * ($c + 1);
            $c = 1;
        }
  
        // prime factorizes a number
        $j = (int)($j / $factor[$j]);
    }
  
    // for the last prime factor
    $ans = $ans * ($c + 1);
  
    return $ans;
}
  
// function to find the 
// smallest integer with
// n factors or more.
function smallest($n)
{
    for ($i = 1;; $i++)
  
        // check if no of 
        // factors is more
        // than n or not
        if (calculateNoOFactors($i) >= $n)
            return $i;
}
  
// Driver Code
  
// generate prime factors 
// of number upto 10^6
generatePrimeFactors();
  
$n = 4;
echo smallest($n);
  
// This code is contributed by mits
?>

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Output:

6

Time Complexity: O(log(max(number))) for every computing number we check before getting the answer.



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Striver(underscore)79 at Codechef and codeforces D

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