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625. Minimum Factorization
Description
Given a positive integer num, return the smallest positive integer x whose multiplication of each digit equals num. If there is no answer or the answer is not fit in 32-bit signed integer, return 0.
Example 1:
Input: num = 48 Output: 68
Example 2:
Input: num = 15 Output: 35
Constraints:
1 <= num <= 231 - 1
Solutions
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class Solution { public int smallestFactorization(int num) { if (num < 2) { return num; } long ans = 0, mul = 1; for (int i = 9; i >= 2; --i) { if (num % i == 0) { while (num % i == 0) { num /= i; ans = mul * i + ans; mul *= 10; } } } return num < 2 && ans <= Integer.MAX_VALUE ? (int) ans : 0; } } -
class Solution { public: int smallestFactorization(int num) { if (num < 2) { return num; } long long ans = 0, mul = 1; for (int i = 9; i >= 2; --i) { if (num % i == 0) { while (num % i == 0) { num /= i; ans = mul * i + ans; mul *= 10; } } } return num < 2 && ans <= INT_MAX ? ans : 0; } }; -
class Solution: def smallestFactorization(self, num: int) -> int: if num < 2: return num ans, mul = 0, 1 for i in range(9, 1, -1): while num % i == 0: num //= i ans = mul * i + ans mul *= 10 return ans if num < 2 and ans <= 2**31 - 1 else 0 -
func smallestFactorization(num int) int { if num < 2 { return num } ans, mul := 0, 1 for i := 9; i >= 2; i-- { if num%i == 0 { for num%i == 0 { num /= i ans = mul*i + ans mul *= 10 } } } if num < 2 && ans <= math.MaxInt32 { return ans } return 0 }