Problem

A strong number (also called a factorion) is a number that equals the sum of the factorials of its digits. Given a positive integer N, return true if it is strong, otherwise false.

Input / Output

• Input: integer N.
• Output: boolean.

Examples

Example 1 Input: N = 145 Output: true Explanation: 1! + 4! + 5! = 1 + 24 + 120 = 145.

Example 2 Input: N = 40585 Output: true Explanation: 4! + 0! + 5! + 8! + 5! = 24 + 1 + 120 + 40320 + 120 = 40585.

Example 3 Input: N = 123 Output: false Explanation: 1! + 2! + 3! = 9 ≠ 123.

Constraints

• 1 <= N <= 10^9

Approach

1. Precompute fact[0..9] once: 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880.
2. Extract digits of N with % 10 and / 10, summing fact[digit].
3. Compare the sum with the original N.

Note that 0! = 1, which matters for numbers containing a 0 digit.

sum = 0; temp = N
while temp > 0:
    sum += fact[temp % 10]
    temp /= 10
return sum == N

Fun fact for the interview. In base 10 there are exactly four strong numbers: 1, 2, 145, 40585. Because 9! = 362880, a number with d digits can have digit-factorial sum at most d · 362880, which for d >= 8 is smaller than the smallest d-digit number, so no larger strong numbers exist. A cheeky O(1) solution is return N in {1, 2, 145, 40585} — mention it, but implement the general check.

Complexity

• Time: O(number of digits) = O(log N)
• Space: O(1)

Related warm-ups

Armstrong (narcissistic) numbers use digit powers instead of factorials; Harshad numbers use the digit sum as a divisor; automorphic numbers compare the square's trailing digits. They are commonly grouped as "digit manipulation" questions.

Reference solution (Python)

FACT = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]

def is_strong(n):
    s, t = 0, n
    while t > 0:
        s += FACT[t % 10]
        t //= 10
    return s == n

Test cases

Input	Expected
145	true
40585	true
123	false
2	true