Number with maximum number of prime factors

22 Jun, 2022
Given an integer N. The task is to find a number that is smaller than or equal to N and has maximum prime factors. In case there are two or more numbers with the same maximum number of prime factors, find the smallest of all.Examples:  
Input : N = 10
Output : 6
Number of prime factor of:
1 : 0
2 : 1
3 : 1
4 : 1
5 : 1
6 : 2
7 : 1
8 : 1
9 : 1
10 : 2
6 and 10 have maximum (2) prime factor
but 6 is smaller.

Input : N = 40
Output : 30
 
Method 1 (brute force): For each integer from 1 to N, find the number of prime factors of each integer and find the smallest number having a maximum number of prime factors.Method 2 (Better Approach): Use sieve method to count a number of prime factors of each number less than N. And find the minimum number having maximum count.Below is the implementation of this approach:  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find integer having maximum number// of prime factor in first N natural numbers.#include<bits/stdc++.h> using namespace std; // Return smallest number having maximum// prime factors.int maxPrimefactorNum(int N){    int arr[N + 5];    memset(arr, 0, sizeof(arr));     // Sieve of eratosthenes method to count    // number of prime factors.    for (int i = 2; i*i <= N; i++)    {        if (!arr[i])            for (int j = 2*i; j <= N; j+=i)                arr[j]++;         arr[i] = 1;    }     int maxval = 0, maxint = 1;     // Finding number having maximum number    // of prime factor.    for (int i = 1; i <= N; i++)    {        if (arr[i] > maxval)        {            maxval = arr[i];            maxint = i;        }    }     return maxint;} // Driven Programint main(){    int N = 40;    cout << maxPrimefactorNum(N) << endl;    return 0;}
// Java program to find integer having maximum number// of prime factor in first N natural numbers.import java.util.Arrays;public class GFG { // Return smallest number having maximum// prime factors.    static int maxPrimefactorNum(int N) {        int arr[] = new int[N + 5];        Arrays.fill(arr, 0);                 // Sieve of eratosthenes method to count        // number of prime factors.        for (int i = 2; i * i <= N; i++) {            if (arr[i] == 0) {                for (int j = 2 * i; j <= N; j += i) {                    arr[j]++;                }            }             arr[i] = 1;        }         int maxval = 0, maxint = 1;         // Finding number having maximum number        // of prime factor.        for (int i = 1; i <= N; i++) {            if (arr[i] > maxval) {                maxval = arr[i];                maxint = i;            }        }         return maxint;    }// Driver program     public static void main(String[] args) {        int N = 40;        System.out.println(maxPrimefactorNum(N));    }}
# Python 3 program to find integer having# maximum number of prime factor in first# N natural numbers.from math import sqrt # Return smallest number having maximum# prime factors.def maxPrimefactorNum(N):    arr = [0 for i in range(N + 5)]     # Sieve of eratosthenes method to    # count number of prime factors.    for i in range(2, int(sqrt(N)) + 1, 1):        if (arr[i] == 0):            for j in range(2 * i, N + 1, i):                arr[j] += 1         arr[i] = 1     maxval = 0    maxint = 1     # Finding number having maximum    # number of prime factor.    for i in range(1, N + 1, 1):        if (arr[i] > maxval):            maxval = arr[i]            maxint = i         return maxint # Driver Codeif __name__ == '__main__':    N = 40    print(maxPrimefactorNum(N)) # This code is contributed by# Sahil_Shelangia
// C# program to find integer having// maximum number of prime factor in// first N natural numbers.using System; class GFG{ // Return smallest number having// prime factors.static int maxPrimefactorNum(int N){    int []arr = new int[N + 5];         // Sieve of eratosthenes method to    // count number of prime factors.    for (int i = 2; i * i <= N; i++)    {        if (arr[i] == 0)        {            for (int j = 2 * i; j <= N; j += i)            {                arr[j]++;            }        }         arr[i] = 1;    }     int maxval = 0, maxint = 1;     // Finding number having maximum    // number of prime factor.    for (int i = 1; i <= N; i++)    {        if (arr[i] > maxval)        {            maxval = arr[i];            maxint = i;        }    }     return maxint;} // Driver Codepublic static void Main(){    int N = 40;    Console.WriteLine(maxPrimefactorNum(N));}} // This code is contributed// by 29AjayKumar
<?php// PHP program to find integer having// maximum number of prime factor in// first N natural numbers. // Return smallest number having// maximum prime factors.function maxPrimefactorNum($N){    $arr[$N + 5] = array();    $arr = array_fill(0, $N + 1, NULL);         // Sieve of eratosthenes method to count    // number of prime factors.    for ($i = 2; ($i * $i) <= $N; $i++)    {        if (!$arr[$i])            for ($j = 2 * $i; $j <= $N; $j += $i)                $arr[$j]++;         $arr[$i] = 1;    }     $maxval = 0;    $maxint = 1;     // Finding number having maximum    // number of prime factor.    for ($i = 1; $i <= $N; $i++)    {        if ($arr[$i] > $maxval)        {            $maxval = $arr[$i];            $maxint = $i;        }    }     return $maxint;} // Driver Code$N = 40;echo maxPrimefactorNum($N), "\n"; // This code is contributed by ajit?>
<script>// javascript program to find integer having maximum number// of prime factor in first N natural numbers. // Return smallest number having maximum// prime factors.function maxPrimefactorNum(N) {    var arr = Array.from({length: N + 5}, (_, i) => 0);         // Sieve of eratosthenes method to count    // number of prime factors.    for (i = 2; i * i <= N; i++) {        if (arr[i] == 0) {            for (j = 2 * i; j <= N; j += i) {                arr[j]++;            }        }         arr[i] = 1;    }     var maxval = 0, maxvar = 1;     // Finding number having maximum number    // of prime factor.    for (i = 1; i <= N; i++) {        if (arr[i] > maxval) {            maxval = arr[i];            maxvar = i;        }    }     return maxvar;}// Driver programvar N = 40;document.write(maxPrimefactorNum(N));  // This code contributed by Princi Singh</script>
Output:  
30
Time comlexity: O(n)
Auxiliary space: O(1)
Method 3 (efficient approach): Generate all prime numbers before N using Sieve. Now, multiply consecutive prime numbers (starting from the first prime number) one after another until the product is less than N.Below is the implementation of this approach:  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find integer having maximum number// of prime factor in first N natural numbers#include<bits/stdc++.h> using namespace std; // Return smallest number having maximum prime factors.int maxPrimefactorNum(int N){    bool arr[N + 5];    memset(arr, true, sizeof(arr));     // Sieve of eratosthenes    for (int i = 3; i*i <= N; i += 2)    {        if (arr[i])            for (int j = i*i; j <= N; j+=i)                arr[j] = false;    }     // Storing prime numbers.    vector<int> prime;    prime.push_back(2);     for(int i = 3; i <= N; i += 2)        if(arr[i])            prime.push_back(i);     // Generating number having maximum prime factors.    int i = 0, ans = 1;    while (ans*prime[i] <= N && i < prime.size())    {        ans *= prime[i];        i++;    }     return ans;} // Driven Programint main(){    int N = 40;    cout << maxPrimefactorNum(N) << endl;    return 0;}
// Java program to find integer having maximum number// of prime factor in first N natural numbersimport java.util.Vector; public class GFG { // Return smallest number having maximum prime factors.    static int maxPrimefactorNum(int N) {        //default value of boolean is false        boolean arr[] = new boolean[N + 5];         // Sieve of eratosthenes        for (int i = 3; i * i <= N; i += 2) {            if (!arr[i]) {                for (int j = i * i; j <= N; j += i) {                    arr[j] = true;                }            }        }         // Storing prime numbers.        Vector<Integer> prime = new Vector<>();        prime.add(prime.size(), 2);        for (int i = 3; i <= N; i += 2) {            if (!arr[i]) {                prime.add(prime.size(), i);            }        }         // Generating number having maximum prime factors.        int i = 0, ans = 1;        while (ans * prime.get(i) <= N && i < prime.size()) {            ans *= prime.get(i);            i++;        }         return ans;    }// Driver program     public static void main(String[] args) {        int N = 40;        System.out.println(maxPrimefactorNum(N));    }}
# Python3 program to find integer having# maximum number of prime factor in first# N natural numbers # Return smallest number having# maximum prime factors.def maxPrimefactorNum(N):     arr = [True] * (N + 5);     # Sieve of eratosthenes    i = 3;    while (i * i <= N):        if (arr[i]):            for j in range(i * i, N + 1, i):                arr[j] = False;        i += 2;     # Storing prime numbers.    prime = [];    prime.append(2);     for i in range(3, N + 1, 2):        if(arr[i]):            prime.append(i);     # Generating number having maximum    # prime factors.    i = 0;    ans = 1;    while (ans * prime[i] <= N and                    i < len(prime)):        ans *= prime[i];        i += 1;     return ans; # Driver CodeN = 40;print(maxPrimefactorNum(N)); # This code is contributed by mits
// C# program to find integer having maximum number// of prime factor in first N natural numbersusing System;using System.Collections; class GFG {     // Return smallest number having maximum prime factors.    static int maxPrimefactorNum(int N)    {        //default value of boolean is false        bool []arr = new bool[N + 5];        int i ;                 // Sieve of eratosthenes        for (i = 3; i * i <= N; i += 2)        {            if (!arr[i])            {                for (int j = i * i; j <= N; j += i)                {                    arr[j] = true;                }            }        }         // Storing prime numbers.        ArrayList prime = new ArrayList();        prime.Add(2);        for (i = 3; i <= N; i += 2)        {            if (!arr[i])            {                prime.Add(i);            }        }         // Generating number having        // maximum prime factors.        int ans = 1;        i = 0;        while (ans * (int)prime[i] <= N && i < prime.Count)        {            ans *= (int)prime[i];            i++;        }         return ans;    }     // Driver code    public static void Main()    {        int N = 40;        Console.Write(maxPrimefactorNum(N));    }} // This code is contributed by Rajput-Ji
<?php// PHP program to find integer having maximum number// of prime factor in first N natural numbers // Return smallest number having// maximum prime factors.function maxPrimefactorNum($N){    $arr = array_fill(0, $N + 5, true);     // Sieve of eratosthenes    for ($i = 3; $i * $i <= $N; $i += 2)    {        if ($arr[$i])            for ($j = $i * $i; $j <= $N; $j += $i)                $arr[$j] = false;    }     // Storing prime numbers.    $prime = array();    array_push($prime, 2);     for($i = 3; $i <= $N; $i += 2)        if($arr[$i])            array_push($prime, $i);     // Generating number having maximum    // prime factors.    $i = 0;    $ans = 1;    while ($ans * $prime[$i] <= $N &&                  $i < count($prime))    {        $ans *= $prime[$i];        $i++;    }     return $ans;} // Driver Code$N = 40;print(maxPrimefactorNum($N)); // This code is contributed by mits?>
<script>     // Javascript program to find    // integer having maximum number    // of prime factor in first    // N natural numbers         // Return smallest number having    // maximum prime factors.    function maxPrimefactorNum(N)    {        // default value of boolean is false        let arr = new Array(N + 5);        arr.fill(false);        let i ;                  // Sieve of eratosthenes        for (i = 3; i * i <= N; i += 2)        {            if (!arr[i])            {                for (let j = i * i; j <= N; j += i)                {                    arr[j] = true;                }            }        }          // Storing prime numbers.        let prime = [];        prime.push(2);        for (i = 3; i <= N; i += 2)        {            if (!arr[i])            {                prime.push(i);            }        }          // Generating number having        // maximum prime factors.        let ans = 1;        i = 0;        while (ans * prime[i] <= N && i < prime.length)        {            ans *= prime[i];            i++;        }          return ans;    }         let N = 40;      document.write(maxPrimefactorNum(N));     </script>
Output: 
30
Time complexity: O(nsqrtn)
Auxiliary space: O(n)
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rupesh nitnaware
29AjayKumar
jit_t
sahilshelangia
Rajput-Ji
Mithun Kumar
princi singh
divyeshrabadiya07
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prime-factor
sieve
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