Program to check Strong Number
Last Updated :
23 Jul, 2025
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++
// C++ program to check if a number is
// strong or not.
#include <bits/stdc++.h>
using namespace std;
int f[10];
// Fills factorials of digits from 0 to 9.
void preCompute()
{
f[0] = f[1] = 1;
for (int i = 2; i<10; ++i)
f[i] = f[i-1] * i;
}
// Returns true if x is Strong
bool isStrong(int x)
{
int factSum = 0;
// Traverse through all digits of x.
int temp = x;
while (temp)
{
factSum += f[temp%10];
temp /= 10;
}
return (factSum == x);
}
// Driver code
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 program to check if
// a number is Strong or not
class CheckStrong
{
static int f[] = new int[10];
// Fills factorials of digits from 0 to 9.
static void preCompute()
{
f[0] = f[1] = 1;
for (int i = 2; i<10; ++i)
f[i] = f[i-1] * i;
}
// Returns true if x is Strong
static boolean isStrong(int x)
{
int factSum = 0;
// Traverse through all digits of x.
int temp = x;
while (temp>0)
{
factSum += f[temp%10];
temp /= 10;
}
return (factSum == x);
}
// main function
public static void main (String[] args)
{
// calling preCompute
preCompute();
// first pass
int x = 145;
if(isStrong(x))
{
System.out.println("Yes");
}
else
System.out.println("No");
// second pass
x = 534;
if(isStrong(x))
{
System.out.println("Yes");
}
else
System.out.println("No");
}
}
# Python program to check if a number is
# strong or not.
f = [None] * 10
# Fills factorials of digits from 0 to 9.
def preCompute() :
f[0] = f[1] = 1;
for i in range(2,10) :
f[i] = f[i-1] * i
# Returns true if x is Strong
def isStrong(x) :
factSum = 0
# Traverse through all digits of x.
temp = x
while (temp) :
factSum = factSum + f[temp % 10]
temp = temp // 10
return (factSum == x)
# Driver code
preCompute()
x = 145
if(isStrong(x) ) :
print ("Yes")
else :
print ("No")
x = 534
if(isStrong(x)) :
print ("Yes")
else:
print ("No")
# This code is contributed by Nikita Tiwari.
// C# program to check if
// a number is Strong or not
using System;
class CheckStrong
{
static int []f = new int[10];
// Fills factorials of digits from 0 to 9.
static void preCompute()
{
f[0] = f[1] = 1;
for (int i = 2; i < 10; ++i)
f[i] = f[i - 1] * i;
}
// Returns true if x is Strong
static bool isStrong(int x)
{
int factSum = 0;
// Traverse through all digits of x.
int temp = x;
while (temp > 0)
{
factSum += f[temp % 10];
temp /= 10;
}
return (factSum == x);
}
// Driver Code
public static void Main ()
{
// calling preCompute
preCompute();
// first pass
int x = 145;
if(isStrong(x))
{
Console.WriteLine("Yes");
}
else
Console.WriteLine("No");
// second pass
x = 534;
if(isStrong(x))
{
Console.WriteLine("Yes");
}
else
Console.WriteLine("No");
}
}
// This code is contributed by Nitin Mittal.
<script>
// Javascript program to check if a number is
// strong or not.
let f = new Array(10);
// Fills factorials of digits from 0 to 9.
function preCompute()
{
f[0] = f[1] = 1;
for (let i = 2; i<10; ++i)
f[i] = f[i-1] * i;
}
// Returns true if x is Strong
function isStrong(x)
{
let factSum = 0;
// Traverse through all digits of x.
let temp = x;
while (temp)
{
factSum += f[temp%10];
temp = Math.floor(temp/10);
}
return (factSum == x);
}
// Driver code
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>");
//This code is contributed by Mayank Tyagi
</script>
<?php
// PHP program to check if a number
// is strong or not.
$f[10] = array();
// Fills factorials of digits
// from 0 to 9.
function preCompute()
{
global $f;
$f[0] = $f[1] = 1;
for ($i = 2; $i < 10; ++$i)
$f[$i] = $f[$i - 1] * $i;
}
// Returns true if x is Strong
function isStrong($x)
{
global $f;
$factSum = 0;
// Traverse through all digits of x.
$temp = $x;
while ($temp)
{
$factSum += $f[$temp % 10];
$temp = (int)$temp / 10;
}
return ($factSum == $x);
}
// Driver code
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";
// This code is contributed by jit_t
?>
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++
// C++ code addition
#include <iostream>
#include <vector>
#include <string>
using namespace std;
string is_strong(int n) {
// Convert the number to an array of digits
vector<int> digits;
int temp = n;
// Loop through each digit
while (temp != 0) {
digits.insert(digits.begin(), temp % 10);
temp /= 10;
}
int factorial_sum = 0;
// Calculate the factorial of the digit
for (int d : digits) {
int f = 1;
for (int i = 1; i <= d; i++) {
f *= i;
}
// Add the factorial to the sum
factorial_sum += f;
}
// Check if the sum of factorials is equal to the original number
if (factorial_sum == n) {
return "Yes";
} else {
return "No";
}
}
int main() {
int n = 145;
cout << is_strong(n) << endl;
return 0;
}
// The code is contributed by Arushi Goel.
import java.util.ArrayList;
public class Main {
// Function to check if a number is a strong number
static String isStrong(int n)
{
// Convert the number to an array of digits
ArrayList<Integer> digits = new ArrayList<>();
int temp = n;
// Loop through each digit
while (temp != 0) {
digits.add(0, temp % 10);
temp /= 10;
}
int factorialSum = 0;
// Calculate the factorial of each digit and add to
// the sum
for (int d : digits) {
int f = 1;
for (int i = 1; i <= d; i++) {
f *= i;
}
factorialSum += f;
}
// Check if the sum of factorials is equal to the
// original number
if (factorialSum == n) {
return "Yes";
}
else {
return "No";
}
}
// Driver's code
public static void main(String[] args)
{
// Test case
int n = 145;
// Displaying whether the number is strong or not
System.out.println(isStrong(n));
}
}
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))
using System;
class Program
{
static string IsStrong(int n)
{
// Convert the number to an array of digits
char[] digits = n.ToString().ToCharArray();
int factorialSum = 0;
// Loop through each digit
foreach (char digit in digits)
{
int d = int.Parse(digit.ToString());
int f = 1;
// Calculate the factorial of the digit
for (int i = 1; i <= d; i++)
{
f *= i;
}
// Add the factorial to the sum
factorialSum += f;
}
// Check if the sum of factorials is equal to the original number
return (factorialSum == n) ? "Yes" : "No";
}
static void Main()
{
int n = 145;
Console.WriteLine(IsStrong(n));
}
}
// Function to check if a number is a strong number
function is_strong(n) {
// Convert the number to an array of digits
let digits = Array.from(String(n), Number);
let factorial_sum = 0;
// Loop through each digit
for (let d of digits) {
let f = 1;
// Calculate the factorial of the digit
for (let i = 1; i <= d; i++) {
f *= i;
}
// Add the factorial to the sum
factorial_sum += f;
}
// Check if the sum of factorials is equal to the original number
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.
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