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:
- 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.
- Then using Greedy Algorithm, we can take largest factorials from this array which can be added to represent N.
- Start from largest possible factorial and keep adding factorials while the remaining value is greater than 0.
- 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++
// C++ program to find minimum number of factorials#include <bits/stdc++.h>#define ll long long intusing 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 14ll fact[14];// Store the actual size of fact[]int size = 1;// Function to precompute factorials till Nvoid 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 Nvoid 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 programint main(){ int n = 27; findMin(n); return 0;} |
Java
// Java program to find minimum number of factorialsimport 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 14static int []fact = new int[14]; // Store the actual size of fact[]static int size = 1; // Function to precompute factorials till Nstatic 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 Nstatic 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 programpublic static void main(String[] args){ int n = 27; findMin(n);}}// This code is contributed by PrinciRaj1992 |
Python3
# 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 14fact = [0]*14# Store the actual size of fact[]size = 1# Function to precompute factorials till Ndef 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 Ndef 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]) # Print min count print(len(ans)) # Print result for i in ans: print(i, end=" ")# Driver programn = 27findMin(n)# This code is contributed by mohit kumar 29 |
C#
// C# program to find minimum number of factorialsusing 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 14static int []fact = new int[14]; // Store the actual size of fact[]static int size = 1; // Function to precompute factorials till Nstatic 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 Nstatic 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 programpublic static void Main(String[] args){ int n = 27; findMin(n);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript 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 14var fact = Array(14).fill(0);// Store the actual size of fact[]var size = 1;// Function to precompute factorials till Nfunction preCompute(N){ // Precomputing factorials fact[0] = 1; for (var 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 Nfunction findMin(N){ // Precompute factorials preCompute(N); var originalN = N; // Initialize result ans = []; // Traverse through all factorials for (var i = size - 1; i >= 0; i--) { // Find factorials while (N >= fact[i]) { N -= fact[i]; ans.push(fact[i]); } } // Print min count document.write(ans.length + "<br>"); // Print result for (var i = 0; i < ans.length; i++) document.write(ans[i] + " ");}// Driver programvar n = 27;findMin(n);// This code is contributed by rutvik_56.</script> |
Output:
3 24 2 1
Time Complexity: O(N * size)
Auxiliary Space: O(N * size)
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