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.

**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; ` `} ` |

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## 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 ` |

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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)

This article is contributed by **Aditya Kamath**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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