What is strong number?

Strong number is a special number whose sum of the factorial of digits is equal to the original number. For Example: 145 is strong number. Since, 1! + 4! + 5!

geeksforgeeks.org - Print all Strong numbers less than or equal to N - GeeksforGeeks

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Print all Strong numbers less than or equal to N

Given a number N, print all the Strong Numbers less than or equal to N. Strong number is a special number whose sum of the factorial of digits is equal to the original number. For Example: 145 is strong number. Since, 1! + 4! + 5! = 145.

Examples:

Input: N = 100

Output: 1 2

Explanation:

Only 1 and 2 are the strong numbers from 1 to 100 because

1! = 1, and

2! = 2 Input: N = 1000

Output: 1 2 145

Explanation:

Only 1, 2 and 145 are the strong numbers from 1 to 1000 because

1! = 1,

2! = 2, and

(1! + 4! + 5!) = 145

Approach: The idea is to iterate from [1, N] and check if any number between the range is strong number or not. If yes then print the corresponding number, else check for the next number.

Below is the implementation of the above approach:

C++

#include using namespace std; int factorial[] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 }; bool isStrong( int N) { string num = to_string(N); int sum = 0; for ( int i = 0; i < num.length(); i++) { sum += factorial[num[i] - '0' ]; } return sum == N; } void printStrongNumbers( int N) { for ( int i = 1; i <= N; i++) { if (isStrong(i)) { cout << i << " " ; } } } int main() { int N = 1000; printStrongNumbers(N); return 0; } Java class GFG { static int [] factorial = { 1 , 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 , 362880 }; public static boolean isStrong( int N) { String num = Integer.toString(N); int sum = 0 ; for ( int i = 0 ; i < num.length(); i++) { sum += factorial[Integer .parseInt(num .charAt(i) + "" )]; } return sum == N; } public static void printStrongNumbers( int N) { for ( int i = 1 ; i <= N; i++) { if (isStrong(i)) { System.out.print(i + " " ); } } } public static void main(String[] args) throws java.lang.Exception { int N = 1000 ; printStrongNumbers(N); } } Python3 factorial = [ 1 , 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 , 362880 ] def isStrong(N): num = str (N) sum = 0 for i in range ( len (num)): sum + = factorial[ ord (num[i]) - ord ( '0' )] if sum = = N: return True else : return False def printStrongNumbers(N): for i in range ( 1 , N + 1 ): if (isStrong(i)): print (i, end = " " ) if __name__ = = "__main__" : N = 1000 printStrongNumbers(N) C#

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using System; class GFG{ static int [] factorial = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 }; public static bool isStrong( int N) { String num = N.ToString(); int sum = 0; for ( int i = 0; i < num.Length; i++) { sum += factorial[ int .Parse(num[i] + "" )]; } return sum == N; } public static void printStrongNumbers( int N) { for ( int i = 1; i <= N; i++) { if (isStrong(i)) { Console.Write(i + " " ); } } } public static void Main(String[] args) { int N = 1000; printStrongNumbers(N); } } Javascript

Output: 1 2 145

Time Complexity: O(N log 10 N)

Auxiliary Space: O(log 10 N)

geeksforgeeks.org - Print all Strong numbers less than or equal to N - GeeksforGeeks

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