Find minimum number X such that sum of factorial of its digits is N
Last Updated :
30 Jan, 2023
Given a number N, the task is to find the minimum number X such that A(X) = N, where A(X) for positive integer X is the sum of factorials of its digits. For example, A(154) = 1! + 5! + 4!= 145. Return a list of digits which represent the number X.
Example:
Input: N = 40321
Output: 18
Explanation: A(18) =1! + 8! = 40321. Note that A(80) and A(81) are also 40321, But 18 is the smallest number.
Input: N = 5040
Output: 7
Approach: To solve the problem follow the below idea:
For any digit n, n! can be written as n * ((n – 1)!). So, digit n – 1 is required exactly n times. This means that taking smaller numbers when a larger number can be taken will just increase the number of digits in our final answer. So, use the greedy approach and subtract the largest value of n! from N until N becomes 0.
Follow the steps to solve this problem:
- Make a factorial array and put all factorials ranging from 0 to 9 in it.
- Keep a variable i and initialize it to 9.
- Now run a while loop and check if factorial[i] <= N.
- Now run another while loop and check how many times you can put subtract factorial[i] from N. Do this for i from 9 to 1.
- Now simply reverse the vector to get the smallest possible number.
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
vector<int> fact;
vector<int> ans;
int helper(int N)
{
if (N == 0)
return 1;
else if (N < 0)
return 0;
for (int i = 9; i >= 0; i--) {
if (fact[i] > N)
continue;
ans.push_back(i);
int d = helper(N - fact[i]);
if (d == 1)
return 1;
ans.erase(find(ans.begin(), ans.end(), i));
}
return 0;
}
vector<int> FactDigit(int N)
{
fact.assign(10, 1);
fact[0] = fact[1] = 1;
for (int i = 2; i <= 9; i++)
fact[i] = i * fact[i - 1];
helper(N);
sort(ans.begin(), ans.end());
return ans;
}
int main()
{
int N = 40321;
vector<int> ans = FactDigit(N);
for (auto i : ans)
cout << i;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static List<Integer> fact = new ArrayList<Integer>(10);
static List<Integer> ans = new ArrayList<Integer>();
static int helper(int N)
{
if (N == 0)
return 1;
else if (N < 0)
return 0;
for (int i = 9; i >= 0; i--) {
if (fact.get(i) > N)
continue;
ans.add(i);
int d = helper(N - fact.get(i));
if (d == 1)
return 1;
ans.remove(ans.get(i));
}
return 0;
}
static List<Integer> FactDigit(int N)
{
fact.add(1);
fact.add(1);
for (int i = 2; i <= 9; i++) {
fact.add(i * fact.get(i - 1));
}
helper(N);
Collections.sort(ans);
return ans;
}
public static void main(String[] args)
{
int N = 40321;
List<Integer> ans = FactDigit(N);
for (int i = 0; i < ans.size(); i++) {
System.out.print(ans.get(i));
}
}
}
|
Python3
fact = []
ans = []
def helper(n)->int:
if(n == 0):
return 1
elif(n<0):
return 0;
i = 9
while(i>=0):
if(fact[i] > n):
i -= 1
continue
ans.append(i)
d = helper(n-fact[i])
if(d==1):
return 1
while(i in ans):
ans.remove(k)
i -= 1
return 0
def FactDigit(n):
fact.append(1)
fact.append(1)
for i in range(2,10):
fact.append(i*fact[i-1])
helper(n)
ans.sort()
return ans
if __name__ == '__main__':
n = 40321
answer = FactDigit(n)
for i in answer:
print(i, end='')
|
C#
using System;
using System.Collections.Generic;
class GFG {
static List<int> fact = new List<int>(10);
static List<int> ans = new List<int>();
static int helper(int N)
{
if (N == 0)
return 1;
else if (N < 0)
return 0;
for (int i = 9; i >= 0; i--) {
if (fact[i] > N)
continue;
ans.Add(i);
int d = helper(N - fact[i]);
if (d == 1)
return 1;
ans.Remove(ans[i]);
}
return 0;
}
static List<int> FactDigit(int N)
{
fact.Add(1);
fact.Add(1);
for (int i = 2; i <= 9; i++) {
fact.Add(i * fact[i - 1]);
}
helper(N);
ans.Sort();
return ans;
}
public static void Main()
{
int N = 40321;
List<int> ans = FactDigit(N);
for (int i = 0; i < ans.Count; i++) {
Console.Write(ans[i]);
}
}
}
|
Javascript
const fact = [];
const ans = [];
function helper(n) {
if (n === 0) {
return 1;
} else if (n < 0) {
return 0;
}
let i = 9;
while (i >= 0) {
if (fact[i] > n) {
i -= 1;
continue;
}
ans.push(i);
const d = helper(n - fact[i]);
if (d === 1) {
return 1;
}
while (ans.includes(i)) {
ans.splice(ans.indexOf(i), 1);
}
i -= 1;
}
return 0;
}
function factDigit(n) {
fact.push(1);
fact.push(1);
for (let i = 2; i < 10; i++) {
fact.push(i * fact[i - 1]);
}
helper(n);
ans.sort();
return ans;
}
const n = 40321;
const answer = factDigit(n);
console.log(answer.join(''));
|
Time Complexity: O(log(N))
Auxiliary Space: O(1), since we are only making a single factorial vector of size 10
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