Program to check Strong Number
Strong Numbers are the numbers whose sum of factorial of digits is equal to the original number. Given a number, check if it is a Strong Number or not.
Examples:
Input : n = 145
Output : Yes
Sum of digit factorials = 1! + 4! + 5!
= 1 + 24 + 120
= 145
Input : n = 534
Output : No
1) Initialize sum of factorials as 0.
2) For every digit d, do following
a) Add d! to sum of factorials.
3) If sum factorials is same as given
number, return true.
4) Else return false.
An optimization is to precompute factorials of all numbers from 0 to 10.
C++
#include <bits/stdc++.h>
using namespace std;
int f[10];
void preCompute()
{
f[0] = f[1] = 1;
for ( int i = 2; i<10; ++i)
f[i] = f[i-1] * i;
}
bool isStrong( int x)
{
int factSum = 0;
int temp = x;
while (temp)
{
factSum += f[temp%10];
temp /= 10;
}
return (factSum == x);
}
int main()
{
preCompute();
int x = 145;
isStrong(x) ? cout << "Yes\n" : cout << "No\n" ;
x = 534;
isStrong(x) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
}
|
Java
class CheckStrong
{
static int f[] = new int [ 10 ];
static void preCompute()
{
f[ 0 ] = f[ 1 ] = 1 ;
for ( int i = 2 ; i< 10 ; ++i)
f[i] = f[i- 1 ] * i;
}
static boolean isStrong( int x)
{
int factSum = 0 ;
int temp = x;
while (temp> 0 )
{
factSum += f[temp% 10 ];
temp /= 10 ;
}
return (factSum == x);
}
public static void main (String[] args)
{
preCompute();
int x = 145 ;
if (isStrong(x))
{
System.out.println( "Yes" );
}
else
System.out.println( "No" );
x = 534 ;
if (isStrong(x))
{
System.out.println( "Yes" );
}
else
System.out.println( "No" );
}
}
|
Python3
f = [ None ] * 10
def preCompute() :
f[ 0 ] = f[ 1 ] = 1 ;
for i in range ( 2 , 10 ) :
f[i] = f[i - 1 ] * i
def isStrong(x) :
factSum = 0
temp = x
while (temp) :
factSum = factSum + f[temp % 10 ]
temp = temp / / 10
return (factSum = = x)
preCompute()
x = 145
if (isStrong(x) ) :
print ( "Yes" )
else :
print ( "No" )
x = 534
if (isStrong(x)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class CheckStrong
{
static int []f = new int [10];
static void preCompute()
{
f[0] = f[1] = 1;
for ( int i = 2; i < 10; ++i)
f[i] = f[i - 1] * i;
}
static bool isStrong( int x)
{
int factSum = 0;
int temp = x;
while (temp > 0)
{
factSum += f[temp % 10];
temp /= 10;
}
return (factSum == x);
}
public static void Main ()
{
preCompute();
int x = 145;
if (isStrong(x))
{
Console.WriteLine( "Yes" );
}
else
Console.WriteLine( "No" );
x = 534;
if (isStrong(x))
{
Console.WriteLine( "Yes" );
}
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
let f = new Array(10);
function preCompute()
{
f[0] = f[1] = 1;
for (let i = 2; i<10; ++i)
f[i] = f[i-1] * i;
}
function isStrong(x)
{
let factSum = 0;
let temp = x;
while (temp)
{
factSum += f[temp%10];
temp = Math.floor(temp/10);
}
return (factSum == x);
}
preCompute();
let x = 145;
isStrong(x) ? document.write( "Yes" + "<br>" ) :
document.write( "No" + "<br>" );
x = 534;
isStrong(x) ? document.write( "Yes" + "<br>" ) :
document.write( "No" + "<br>" );
</script>
|
PHP
<?php
$f [10] = array ();
function preCompute()
{
global $f ;
$f [0] = $f [1] = 1;
for ( $i = 2; $i < 10; ++ $i )
$f [ $i ] = $f [ $i - 1] * $i ;
}
function isStrong( $x )
{
global $f ;
$factSum = 0;
$temp = $x ;
while ( $temp )
{
$factSum += $f [ $temp % 10];
$temp = (int) $temp / 10;
}
return ( $factSum == $x );
}
preCompute();
$x = 145;
if (isStrong(! $x ))
echo "Yes\n" ;
else
echo "No\n" ;
$x = 534;
if (isStrong( $x ))
echo "Yes\n" ;
else
echo "No\n" ;
?>
|
Time Complexity: O(logn)
Auxiliary Space: O(1), since constant extra space has been taken.
Approach#2: Using Iterative Method
This approach converts the input number into a list of its digits. It then computes the factorial sum of each digit and adds them up. If the final sum is equal to the input number, it returns “Yes”, otherwise it returns “No”.
Algorithm
1. Define a function is_strong(n) that takes a number n as input.
2. Convert the number to a string and get its digits.
3. Calculate the factorial of each digit using an iterative method.
4. Sum the factorials of all digits.
5. If the sum is equal to the number n, return “Yes”, otherwise return “No”.
C++
#include <iostream>
#include <vector>
#include <string>
using namespace std;
string is_strong( int n) {
vector< int > digits;
int temp = n;
while (temp != 0) {
digits.insert(digits.begin(), temp % 10);
temp /= 10;
}
int factorial_sum = 0;
for ( int d : digits) {
int f = 1;
for ( int i = 1; i <= d; i++) {
f *= i;
}
factorial_sum += f;
}
if (factorial_sum == n) {
return "Yes" ;
} else {
return "No" ;
}
}
int main() {
int n = 145;
cout << is_strong(n) << endl;
return 0;
}
|
Java
import java.util.ArrayList;
public class Main {
static String isStrong( int n)
{
ArrayList<Integer> digits = new ArrayList<>();
int temp = n;
while (temp != 0 ) {
digits.add( 0 , temp % 10 );
temp /= 10 ;
}
int factorialSum = 0 ;
for ( int d : digits) {
int f = 1 ;
for ( int i = 1 ; i <= d; i++) {
f *= i;
}
factorialSum += f;
}
if (factorialSum == n) {
return "Yes" ;
}
else {
return "No" ;
}
}
public static void main(String[] args)
{
int n = 145 ;
System.out.println(isStrong(n));
}
}
|
Python3
def is_strong(n):
digits = [ int (d) for d in str (n)]
factorial_sum = 0
for d in digits:
f = 1
for i in range ( 1 , d + 1 ):
f * = i
factorial_sum + = f
if factorial_sum = = n:
return "Yes"
else :
return "No"
n = 145
print (is_strong(n))
|
C#
using System;
class Program
{
static string IsStrong( int n)
{
char [] digits = n.ToString().ToCharArray();
int factorialSum = 0;
foreach ( char digit in digits)
{
int d = int .Parse(digit.ToString());
int f = 1;
for ( int i = 1; i <= d; i++)
{
f *= i;
}
factorialSum += f;
}
return (factorialSum == n) ? "Yes" : "No" ;
}
static void Main()
{
int n = 145;
Console.WriteLine(IsStrong(n));
}
}
|
Javascript
function is_strong(n) {
let digits = Array.from(String(n), Number);
let factorial_sum = 0;
for (let d of digits) {
let f = 1;
for (let i = 1; i <= d; i++) {
f *= i;
}
factorial_sum += f;
}
if (factorial_sum == n) {
return "Yes" ;
} else {
return "No" ;
}
}
let n = 145;
console.log(is_strong(n));
|
Time Complexity: O(kn^2), where k is the number of digits in the number and n is the maximum value of a digit
Auxiliary Space: O(k), where k is the number of digits in the number.
Last Updated :
18 Dec, 2023
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