Smallest number S such that N is a factor of S factorial or S!

• Difficulty Level : Medium
• Last Updated : 01 Feb, 2021

Given a number N. You are tasked with finding the smallest number S, such that N is a factor of S! (S factorial). N can be very large.
Examples:

Input  : 6
Output : 3
The value of 3! is 6
This is the smallest number which can have 6 as a factor.

Input  : 997587429953
Output : 998957
If we calculate out 998957!,
we shall find that it is divisible by 997587429953.
Factors of 997587429953 are 998957 and 998629.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Naive Approach
We iterate from 1 to N, calculating factorial in each case. When we find a factorial that’s capable of having N as a factor, we output it. This method will be difficult to implement for large N, as the factorial can become very large.
Time Complexity: O(N^2)
Optimized Naive Approach
Instead of iterating from 1 to N, we use binary search. This is still a bad method, as we are still trying to calculate N!
Time Complexity O(N log N)
Optimum Solution
We can first calculate all the prime factors of N. We then reduce our problem to finding a factorial which has all the prime factors of N, at least as many times as they appear in N. We then binary search on elements from 1 to N. We can utilize Legendre’s Formula to check whether a number’s factorial has all the same prime factors. We then find the smallest such number.

C++

// Program to find factorial that N belongs to
#include <bits/stdc++.h>
using namespace std;

#define ull unsigned long long

// Calculate prime factors for a given number
map<ull, int> primeFactors(ull num)
{
// Container for prime factors
map<ull, int> ans;

// Iterate from 2 to i^2 finding all factors
for (ull i = 2; i * i <= num; i++)
{
while (num % i == 0)
{
num /= i;
ans[i]++;
}
}

// If we still have a remainder
// it is also a prime factor
if (num > 1)
ans[num]++;
return ans;
}

// Calculate occurrence of an element
// in factorial of a number
ull legendre(ull factor, ull num)
{
ull count = 0, fac2 = factor;
while (num >= factor)
{
count += num / factor;
factor *= fac2;
}
return count;
}

bool possible(map<ull, int> &factors, ull num)
{
// Iterate through prime factors
for (map<ull, int>::iterator it = factors.begin();
it != factors.end(); ++it)
{
// Check if factorial contains less
// occurrences of prime factor
if (legendre(it->first, num) < it->second)
return false;
}
return true;
}

// Function to binary search 1 to N
ull search(ull start, ull end, map<ull, int> &factors)
{
ull mid = (start + end) / 2;

// Prime factors are not in the factorial
// Increase the lowerbound
if (!possible(factors, mid))
return search(mid + 1, end, factors);

// We have reached smallest occurrence
if (start == mid)
return mid;

// Smaller factorial satisfying
// requirements may exist, decrease upperbound
return search(start, mid, factors);
}

// Calculate prime factors and search
ull findFact(ull num)
{
map<ull, int> factors = primeFactors(num);
return search(1, num, factors);
}

// Driver function
int main()
{
cout << findFact(6) << "n";
cout << findFact(997587429953) << "n";
return 0;
}

Java

// Java Program to find factorial that N belongs to

import java.util.HashMap;
import java.util.Iterator;
import java.util.Set;

class Test
{
// Calculate prime factors for a given number
static HashMap<Long, Integer> primeFactors(long num)
{

// Container for prime factors
HashMap<Long, Integer> ans = new HashMap<Long, Integer>(){
@Override
public Integer get(Object key) {
if(containsKey(key)){
return super.get(key);
}
return 0;
}
};

// Iterate from 2 to i^2 finding all factors
for (long i = 2; i * i <= num; i++)
{
while (num % i == 0)
{
num /= i;
ans.put(i, ans.get(i)+1);
}
}

// If we still have a remainder
// it is also a prime factor
if (num > 1)
ans.put(num, ans.get(num)+1);;
return ans;
}

// Calculate occurrence of an element
// in factorial of a number
static long legendre(long factor, long num)
{
long count = 0, fac2 = factor;
while (num >= factor)
{
count += num / factor;
factor *= fac2;
}
return count;
}

static boolean possible(HashMap<Long, Integer> factors, long num)
{
Set<Long> s = factors.keySet();

// Iterate through prime factors
Iterator<Long> itr = s.iterator();

while (itr.hasNext()) {
long temp = itr.next();
// Check if factorial contains less
// occurrences of prime factor
if (legendre(temp, num) < factors.get(temp))
return false;
}

return true;
}

// Method to binary search 1 to N
static long search(long start, long end, HashMap<Long, Integer> factors)
{
long mid = (start + end) / 2;

// Prime factors are not in the factorial
// Increase the lowerbound
if (!possible(factors, mid))
return search(mid + 1, end, factors);

// We have reached smallest occurrence
if (start == mid)
return mid;

// Smaller factorial satisfying
// requirements may exist, decrease upperbound
return search(start, mid, factors);
}

// Calculate prime factors and search
static long findFact(long num)
{
HashMap<Long, Integer>  factors = primeFactors(num);
return search(1, num, factors);
}

// Driver method
public static void main(String args[])
{
System.out.println(findFact(6));
System.out.println(findFact(997587429953L));
}
}
// This code is contributed by Gaurav Miglani

Python3

# Python Program to find factorial that N belongs to

# Calculate prime factors for a given number
def primeFactors(num):

# Container for prime factors
ans = dict()
i = 2

# Iterate from 2 to i^2 finding all factors
while(i * i <= num):
while (num % i == 0):
num //= i;
if i not in ans:
ans[i] = 0
ans[i] += 1

# If we still have a remainder
# it is also a prime factor
if (num > 1):
if num not in ans:
ans[num] = 0
ans[num] += 1
return ans;

# Calculate occurrence of an element
# in factorial of a number
def legendre(factor, num):
count = 0
fac2 = factor;
while (num >= factor):
count += num // factor;
factor *= fac2;
return count;

def possible(factors, num):

# Iterate through prime factors
for it in factors.keys():

# Check if factorial contains less
# occurrences of prime factor
if (legendre(it, num) < factors[it]):
return False;
return True;

# Function to binary search 1 to N
def search(start, end, factors):
mid = (start + end) // 2;

# Prime factors are not in the factorial
# Increase the lowerbound
if (not possible(factors, mid)):
return search(mid + 1, end, factors);

# We have reached smallest occurrence
if (start == mid):
return mid;

# Smaller factorial satisfying
# requirements may exist, decrease upperbound
return search(start, mid, factors);

# Calculate prime factors and search
def findFact(num):
factors = primeFactors(num);
return search(1, num, factors);

# Driver function
if __name__=='__main__':

print(findFact(6))
print(findFact(997587429953))

# This code is contributed by pratham76.

C#

// C# Program to find factorial that N belongs to
using System;
using System.Collections;
using System.Collections.Generic;

class Test
{

// Calculate prime factors for a given number
static Dictionary<long, int> primeFactors(long num)
{

// Container for prime factors
Dictionary<long, int> ans = new Dictionary<long, int>();

// Iterate from 2 to i^2 finding all factors
for (long i = 2; i * i <= num; i++)
{
while (num % i == 0)
{
num /= i;
if(!ans.ContainsKey(i))
{
ans[i] = 0;
}
ans[i]++;
}
}

// If we still have a remainder
// it is also a prime factor
if (num > 1)
{
if(!ans.ContainsKey(num))
{
ans[num] = 0;
}
ans[num]++;
}
return ans;
}

// Calculate occurrence of an element
// in factorial of a number
static long legendre(long factor, long num)
{
long count = 0, fac2 = factor;
while (num >= factor)
{
count += num / factor;
factor *= fac2;
}
return count;
}

static bool possible(Dictionary<long, int> factors, long num)
{

foreach (int itr in factors.Keys)
{
// Check if factorial contains less
// occurrences of prime factor
if (legendre(itr, num) < factors[itr])
return false;
}
return true;
}

// Method to binary search 1 to N
static long search(long start, long end, Dictionary<long, int> factors)
{
long mid = (start + end) / 2;

// Prime factors are not in the factorial
// Increase the lowerbound
if (!possible(factors, mid))
return search(mid + 1, end, factors);

// We have reached smallest occurrence
if (start == mid)
return mid;

// Smaller factorial satisfying
// requirements may exist, decrease upperbound
return search(start, mid, factors);
}

// Calculate prime factors and search
static long findFact(long num)
{
Dictionary<long, int>  factors = primeFactors(num);
return search(1, num, factors);
}

// Driver method
public static void Main()
{
Console.WriteLine(findFact(6));
Console.WriteLine(findFact(997587429953L));
}
}

// This code is contributed by rutvik_56.

Output:

3
998957

At no point do we actually calculate a factorial. This means we do not have to worry about the factorial being too large to store.
Lagrange’s Formula runs in O(Log N).
Binary search is O(Log N).
Calculating prime factors is O(sqrt(N))
Iterating through prime factors is O(Log N).
Time complexity becomes: O(sqrt(N) + (Log N)^3)