Minimum number of Factorials whose sum is equal to N

Given a number N (<1010), the task is to find the minimum number of factorials needed to represent N, as their sum. Also, print those factorials.

Examples:

Input: N = 30
Output: 2
24, 6
Explanation:
Factorials needed to represent 30: 24, 6

Input: N = 150
Output: 3
120, 24, 6
Explanation:
Factorials needed to represent 150: 120 24 6

Approach:

  1. In order to efficiently find the factorials which are needed to represent N as their sum, we can precompute the factorials till N (N < 1010) and store them in an array, for faster calculations.
  2. Then using Greedy Algorithm, we can take largest factorials from this array which can be added to represent N.
  3. Start from largest possible factorial and keep adding factorials while the remaining value is greater than 0.
  4. Below is the complete algorithm.
    • Initialize result as empty
    • find the largest factorial that is smaller than N
    • Add found factorial to result. Subtract value of found factorial from N
    • If N becomes 0, then print result. Else repeat steps 2 and 3 for new value of N

Below is the implementation of the above approach:

C++

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// C++ program to find minimum number of factorials
  
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
  
// Array to calculate all factorials
// less than or equal to N
// Since there can be only 14 factorials
// till 10^10
// Hence the maximum size of fact[] is 14
ll fact[14];
  
// Store the actual size of fact[]
int size = 1;
  
// Function to precompute factorials till N
void preCompute(int N)
{
    // Precomputing factorials
    fact[0] = 1;
  
    for (int i = 1; fact[i - 1] <= N; i++) {
        fact[i] = (fact[i - 1] * i);
        size++;
    }
}
  
// Function to find the minimum number
// of factorials whose sum represents N
void findMin(int N)
{
  
    // Precompute factorials
    preCompute(N);
  
    int originalN = N;
  
    // Initialize result
    vector<int> ans;
  
    // Traverse through all factorials
    for (int i = size - 1; i >= 0; i--) {
  
        // Find factorials
        while (N >= fact[i]) {
            N -= fact[i];
            ans.push_back(fact[i]);
        }
    }
  
    // Print min count
    cout << ans.size() << "\n";
  
    // Print result
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << " ";
}
  
// Driver program
int main()
{
    int n = 27;
    findMin(n);
    return 0;
}

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Java

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// Java program to find minimum number of factorials
import java.util.*;
  
class GFG{
   
// Array to calculate all factorials
// less than or equal to N
// Since there can be only 14 factorials
// till 10^10
// Hence the maximum size of fact[] is 14
static int []fact = new int[14];
   
// Store the actual size of fact[]
static int size = 1;
   
// Function to precompute factorials till N
static void preCompute(int N)
{
    // Precomputing factorials
    fact[0] = 1;
   
    for (int i = 1; fact[i - 1] <= N; i++) {
        fact[i] = (fact[i - 1] * i);
        size++;
    }
}
   
// Function to find the minimum number
// of factorials whose sum represents N
static void findMin(int N)
{
   
    // Precompute factorials
    preCompute(N);
   
    int originalN = N;
   
    // Initialize result
    Vector<Integer> ans = new Vector<Integer>();
   
    // Traverse through all factorials
    for (int i = size - 1; i >= 0; i--) {
   
        // Find factorials
        while (N >= fact[i]) {
            N -= fact[i];
            ans.add(fact[i]);
        }
    }
   
    // Print min count
    System.out.print(ans.size()+ "\n");
   
    // Print result
    for (int i = 0; i < ans.size(); i++)
        System.out.print(ans.get(i)+ " ");
}
   
// Driver program
public static void main(String[] args)
{
    int n = 27;
    findMin(n);
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to find minimum number of factorials
  
# Array to calculate all factorials
# less than or equal to N
# Since there can be only 14 factorials
# till 10^10
# Hence the maximum size of fact[] is 14
fact = [0]*14
  
# Store the actual size of fact[]
size = 1
  
# Function to precompute factorials till N
def preCompute(N):
    global size
      
    # Precomputing factorials
    fact[0] = 1
  
    i = 1
  
    while fact[i - 1] <= N:
        fact[i] = fact[i - 1] * i
        size += 1
        i += 1
  
# Function to find the minimum number
# of factorials whose sum represents N
def findMin(N):
  
    # Precompute factorials
      
    preCompute(N)
  
    originalN = N
  
    # Initialize result
    ans = []
  
    # Traverse through all factorials
    for i in range(size-1, -1, -1):
  
        # Find factorials
        while (N >= fact[i]):
            N -= fact[i]
            ans.append(fact[i])
  
    # Prmin count
    print(len(ans))
  
    # Prresult
    for i in ans:
        print(i, end=" ")
  
# Driver program
  
n = 27
findMin(n)
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to find minimum number of factorials
using System;
using System.Collections.Generic;
  
class GFG{
    
// Array to calculate all factorials
// less than or equal to N
// Since there can be only 14 factorials
// till 10^10
// Hence the maximum size of fact[] is 14
static int []fact = new int[14];
    
// Store the actual size of fact[]
static int size = 1;
    
// Function to precompute factorials till N
static void preCompute(int N)
{
    // Precomputing factorials
    fact[0] = 1;
    
    for (int i = 1; fact[i - 1] <= N; i++) {
        fact[i] = (fact[i - 1] * i);
        size++;
    }
}
    
// Function to find the minimum number
// of factorials whose sum represents N
static void findMin(int N)
{
    
    // Precompute factorials
    preCompute(N);
    
    int originalN = N;
    
    // Initialize result
    List<int> ans = new List<int>();
    
    // Traverse through all factorials
    for (int i = size - 1; i >= 0; i--) {
    
        // Find factorials
        while (N >= fact[i]) {
            N -= fact[i];
            ans.Add(fact[i]);
        }
    }
    
    // Print min count
    Console.Write(ans.Count+ "\n");
    
    // Print result
    for (int i = 0; i < ans.Count; i++)
        Console.Write(ans[i]+ " ");
}
    
// Driver program
public static void Main(String[] args)
{
    int n = 27;
    findMin(n);
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

3
24 2 1

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