In this guide, you will get 1362. Closest Divisors LeetCode Solution with the best time and space complexity. The solution to Closest Divisors problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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
Complexity Analysis
Time Complexity:
Space Complexity:
1362. Closest Divisors LeetCode Solution in C++
class Solution {
public:
vector<int> closestDivisors(int num) {
for (int root = sqrt(num + 2); root > 0; --root)
for (int cand : {num + 1, num + 2})
if (cand % root == 0)
return {root, cand / root};
throw;
}
}; /* code provided by PROGIEZ */
1362. Closest Divisors LeetCode Solution in Java
class Solution {
public int[] closestDivisors(int num) {
for (int root = (int) Math.sqrt(num + 2); root > 0; --root)
for (int cand : new int[] {num + 1, num + 2})
if (cand % root == 0)
return new int[] {root, cand / root};
throw new IllegalArgumentException();
}
} // code provided by PROGIEZ
1362. Closest Divisors LeetCode Solution in Python
class Solution:
def closestDivisors(self, num: int) -> list[int]:
for root in reversed(range(math.isqrt(num + 2) + 1)):
for cand in [num + 1, num + 2]:
if cand % root == 0:
return [root, cand // root] # code by PROGIEZ
