LeetCode 1362: Closest Divisors Solution
Master LeetCode problem 1362 (Closest Divisors), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1362. Closest Divisors
Problem Explanation
Explanation
To solve this problem, we need to find the closest two divisors of num + 1 or num + 2. We can start by iterating from the square root of num + 2 down to 1 to find the divisors. For each divisor found, we calculate the corresponding quotient and check if it is a valid divisor. We keep track of the closest pair of divisors found so far and update it when a closer pair is found.
Solution Code
class Solution {
public int[] closestDivisors(int num) {
int[] result1 = findClosestDivisors(num + 1);
int[] result2 = findClosestDivisors(num + 2);
return Math.abs(result1[0] - result1[1]) < Math.abs(result2[0] - result2[1]) ? result1 : result2;
}
private int[] findClosestDivisors(int n) {
int sqrt = (int) Math.sqrt(n);
for (int i = sqrt; i >= 1; i--) {
if (n % i == 0) {
return new int[] {i, n / i};
}
}
return new int[] {1, n};
}
}Try It Yourself
Loading code editor...
Related LeetCode Problems
Frequently Asked Questions
How to solve LeetCode 1362 (Closest Divisors)?
This page provides optimized solutions for LeetCode problem 1362 (Closest Divisors) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.
What is the time complexity of LeetCode 1362 (Closest Divisors)?
The time complexity for LeetCode 1362 (Closest Divisors) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.
Can I run code for LeetCode 1362 on DevExCode?
Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 1362 in Java, C++, or Python.