Leetcode Problem 1362 Closest Divisors.txt

1362. Closest Divisors
Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.

Return the two integers in any order.

 

Example 1:

Input: num = 8
Output: [3,3]
Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
Example 2:

Input: num = 123
Output: [5,25]
Example 3:

Input: num = 999
Output: [40,25]
 

Constraints:

1 <= num <= 10^9


Hint to solve: Find divisors until the sqrt of the number

class Solution:
    
    def divisors(self,num):
        
        closestDiff = num
        result = [0,0]
        for i in range(1,(int)(math.sqrt(num)+1)):
            if num % i == 0:
                first = i
                second =(int)(num/i)
                diff = abs(first - second)
                if diff < closestDiff:
                    result[0] = first
                    result[1] = second
                    closestDiff = diff
        return result
                
    
    def closestDivisors(self, num: int) -> List[int]:
        
        num1 = num + 1
        firstResult = self.divisors(num1)
        print(f"firstResult: {firstResult}")
        
        num2 = num + 2
        secondResult = self.divisors(num2)
        print(f"secondResult:{secondResult}")
        
        if abs(firstResult[0] - firstResult[1]) < abs(secondResult[0] - secondResult[1]):
            return firstResult
        else:
            return secondResult